782: Analytic Number Theory (Instructor s Notes)*

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1 Math 782: Analytic Number Theory Instructor s Notes)* Analytic Versus Elementary: Terminology Analytic Number Theory makes use of Comple Analysis and Elementary Number Theory does not; but it isn t so simple to distinguish.) Writing an integer as a sum of two squares. This is the first of a few eamples of how Comple Analysis can be used to answer a question seemingly unrelated to it: if a and b are sums of 2 squares, why must ab be also? Coverings. A covering of the integers is a system of congruences a j mod m j ) such that every integer satisfies at least one of these congruences. A perfect covering is one in which each integer satisfies eactly one of the congruences. Suppose we have a finite perfect covering of the integers with each modulus m j >. Show that the largest modulus occurs twice. Use that r z = z a j z m for z <, j j= where <m m 2 m r and a j <m j. Suppose that m r m r. Then the right-hand side, and not the left-hand side, gets large in absolute value) as z approaches ep2πi/m r ). The contradiction implies m r = m r. Preliminaries on eponentials. The following are useful formulas: n { if k e i2πk/n)j = n if k =, j= e i2πkθ dθ = 2π { if k e ikθ dθ = 2π if k =. Putnam Problem A-6, 985. If p) =a +a + +a m m is a polynomial with real coefficients a j, then set Γp)) = a 2 + a a 2 m. Let f) = Find, with proof, a polynomial g) with real coefficients such that i) g) =, and ii) Γf) n )=Γg) n ) for all positive integers n. There are perhaps better ways to do this problem, but we use what we just learned to establish Γp)) = p e 2πiθ) p e 2πiθ) dθ. Note that f) =3+ ) + 2). That g) = =3+ )2 + ) provides a solution follows from Γf) n )= f e 2πiθ) n f e 2πiθ ) n dθ *These notes are from a course taught by Michael Filaseta in the Fall of 996.

2 2 = = = 3e 2πiθ + ) n e 2πiθ +2 ) n 3e 2πiθ + ) n e 2πiθ +2 ) n dθ 3e 2πiθ + ) n 2e 2πiθ + ) n 3e 2πiθ + ) n 2e 2πiθ + ) n dθ g e 2πiθ) n g e 2πiθ ) n dθ =Γg) n ). A Geometric Problem. Let R be a rectangle, and suppose R can be epressed as a union of rectangles R j with edges parallel to R and common points only along these edges. Suppose each R j has at least one edge of integer length. Then R itself must have an edge of integer length. Let R be positioned so that it is in the first quadrant of the y-plane, with an edge on each of the and y aes. Let R j have bottom left-hand corner u j,v j ), and let α j and β j be the horizontal and vertical dimensions of R j respectively. Observe that w+γ w e 2πit dt = e 2πiγ = γ Z. 2π One uses that one of α j and β j is an integer for each j and that r e 2πi+y) d dy = e 2πi+y) d dy to obtain the desired result. R Homework: ) a) If a and b are integers which can each be epressed in the form 2 +5y 2 for some integers and y, eplain why it is possible to epress ab in this form as well. b) Determine whether the following is true: if a and b are integers and ab can be epressed in the form 2 +5y 2 with and y integers, then either a or b can be as well. Either prove the result or give a countereample. 2) Putnam Problem A-5, 985) For m a positive integer, define I m = 2π j= R j cos cos2) cos3) cosm) d. a) Use one of the preliminary results above yes, you must to get credit for the problem) to determine with proof for which integers m with m we have I m. Hint: Use that cos =e i + e i )/2.)

3 b) Generalize part a). In other words, determine which positive integers m are such that I m. The answer should be in the form: I m if and only if m u mod v) where v is an integer and u is a list of integers. Justify your answer. Elementary Aspects: The Fundamental Theorem of Arithmetic the atoms of which the matter called integers is made are primes ) There are infinitely many primes Give three proofs: i) Euclid s, ii) 2 2n + is divisible by a prime > 2 n+, and iii) p /p2 ) = π 2 /6, an irrational number.) Eplain the notion of rearranging the terms of a series as well as the following lemmas about absolute convergence. Lemma. If the terms of an absolutely converging series a n with a n C are rearranged, then the value of the series remains unchanged. On the other hand, if the series a n with a n R converges but not absolutely, then any real value can be obtained from an appropriate rearrangement of the series. Use the eample of the alternating harmonic series )n+ /n to illustrate the second half of the lemma and to give the first half more meaning). For the first half, let S = a n and let a n be a rearrangement of it. Let ε>. Fi n such that n>n a n <ε. Let N be sufficiently large so that the terms a,...,a n appear among a,...,a N. Then N a n S a n <ε. n>n The first part of the lemma follows. Lemma 2. The product of two absolutely converging series converges absolutely. The product of a n and b n is n=2 c n where c n = n k= a kb n k. The result follows since N n=2 c n is an increasing function of N bounded above by N ) N a n ) b n ) a n ) b n. In fact, k= l= a kb l converges. Lemma 2 is of interest but is not really needed in what follows. Homework: ) Let f :Z + ) 2 C. We say that the iterated series m= fm, n) = m= ) fm, n) 3

4 4 converges if for each positive integer n, the series m= fm, n) converges, and if lim N N m= ) fm, n) eists. The value of this limit is called the value of the iterated series. We say that the double series fm, n) n< m< converges if there is an S such that for every ε>, there eists a K = Kε) such that if N and M are positive integers K, then N M m= ) fm, n) S <ε. The number S is called the value of the double series. Suppose the double series is absolutely convergent, in other words that fm, n) converges. Prove that each of n< m< fm, n), m= n<, m< fm, n), converges and that their three values are equal. and m= fm, n) Euler s product identity, namely ) p s = n s p for s>. Let P ) denote the product above restricted to primes p. Use Lemma to show The identity follows. [] n s P ) n s. First proof that p /p diverges. Assume not. Fi N such that p>n /p < /2. Then p N ) + p p + p>n p>n p ) 2 + M n

5 5 for any M>. This is a contradiction the left-hand side is finite). Logarithms and a n). Suppose that a n /2. This condition implies ) k=2 a k n k a k n 2a 2 n a n. k=2 We use Observe that log ) = log k= k k N a n )= for < actually for <). N log a n ) 2 N a n. The left-hand side is decreasing so that if a n converges, then so does a n). The inequality log N a n) N a n implies that if a n) converges, then so does a n. The estimate p Use p ) p log. p) [] n dt = log. t Second proof that p /p diverges. Let P ) = p /p)) and S) = p /p. Then by ) log P ) =S)+E where E 2 p p 2 2. The result follows since P ) log. Note: we have found a good lower bound for p /p.) A noteworthy eample. The product p /p)) and µn)/n are related if you epand the product and rearrange the terms, you get the sum). The value of the product is and the value of the sum is ; however, it is considerably harder to show the latter. It is equivalent to the Prime Number Theorem). The notation π) and a proof that lim π)/ =. Show that in fact π) <C/log log for some constant C by using the sieve of Eratosthenes. Discuss sieves in general.)

6 6 Homework: ) a) Modify the previous argument to show that almost all positive integers n have a prime factor log n. More precisely and moreover) show that {n : n does not have a prime factor log n} is bounded above by C/log log for some constant C. Hint: Most n which do not have a prime factor log n also do not have a prime factor log )/2.) b) Suppose wn) is any function defined for n Z + which tends to infinity with n. Prove that {n : n has a prime factor wn)} lim =. 2) Give the analysis details to the second half of the proof of Lemma. The Functions π), ϑ), and ψ), and Chebyshev s Estimate: Define ϑ) = log p and ψ) = log p = [ ] log log p. log p p p m p The limit connection with possibly infinite limits): Theorem. The following hold: lim inf ϑ) ψ) = lim inf π) log = lim inf and lim sup ϑ) ψ) = lim sup = lim sup π) log. More generally, if n is such that lim n n = and any of the limits lim n ϑ n )/ n, lim n ψ n )/ n, lim n π n ) log n )/ n eists, then they all eist and are equal. Comment: This theorem follows fairly quickly from the fact that lim π)/ = and Abel summation to be discussed momentarily). Proof of Theorem. We show that for every ε> and for sufficiently large ε)), π) log ε) + o) ϑ) ψ) π) log, from which the theorem follows. The second inequality is obvious. The last inequality is a consequence of ψ) = p [ ] log log p log = π) log. log p p

7 7 For the first inequality, suppose as we may that ε< and take w = ε. From ϑ) = log p log p w log )π) π w )), p w <p we deduce that ϑ) and the first inequality follows. π) log w w log w, Chebyshev s Theorem in a weaker form) Theorem. If is sufficiently large, then.69 log <π)<2.78 log. Lemma. Let n be a positive integer, and let p be a prime. Then the non-negative integer k for which p k eactly divides n! is [ ] [ ] [ ] n n n k = + p p 2 + p 3 +. Proof of Theorem. Let N = 2n n ). Let kp denote the integer k such that p k eactly divides N. From the Lemma, we deduce that k p = j= [ ] [ ]) 2n n p j 2 p j. The upper limit of summation could be replaced by [log 2n)/log p)]. Observe that [2] 2[] is or depending respectively on whether =[]+{}is such that {} < /2 or not. It follows that k p [log 2n)/log p)]. Using ψ) = [ ] p log )/log p) log p and that N = p 2n pk p, we deduce that log N ψ2n). Since N is the largest coefficient in +) 2n to see this use that ) m k = m k+ m k k ) ), we deduce that i) N<2 2n and ii) 2n +)N>2 2n. We are interested in ii) for the moment which implies a lower bound on log N. We deduce from the above that For >2 and n =[/2], we obtain ψ2n) 2n log 2 log2n +). ψ) ψ2n) > 2) log 2 log +). Dividing by, we deduce that lim inf ψ)/ log 2 >.69. The previous theorem now implies the lower bound in Chebyshev s theorem.

8 8 The upper bound can be obtained by using n<p 2n p N and i) to deduce that ϑ2n) ϑn) = n<p 2n log p log N<2nlog 2. Let > and denote by m the non-negative integer for which 2 m < 2 m+. Letting n =,2,4,...,2 m above and summing, we obtain ϑ) ϑ2 m+ )= m j= ϑ 2 j+) ϑ 2 j)) < 2 m+2 log 2 < 4 log 2). It follows that lim sup ϑ)/ 4 log 2 < theorem now follows from the previous theorem. The upper bound in Chebyshev s Why did Chebyshev care? Chebyshev was interested in proving Bertrand s hypothesis that for every positive integer n, there is a prime in the interval n, 2n]. He did this by obtaining a stronger version of his theorem than we have stated here. He showed that.92 log <π)<. log for sufficiently large. It follows that for n sufficiently large 2n π2n) πn) >.92 log2n). n log n > which establishes Bertrand s hypothesis when n is sufficiently large). To complete the proof of Bertrand s hypothesis, one can determine what sufficiently large means here and then check the hypothesis directly for the n which are not sufficiently large. Homework: ) a) Using the ideas above and Chebyshev s theorem as we have established it, find a constant c as small as you can such that for every ε>and for sufficiently large depending on ε), there is a prime in the interval, c + ε)]. b) Find a constant c as small as you can such that for every ε>and for sufficiently large depending on ε), there are at least 3 primes in the interval, c + ε)]. 2) a) Show that for n sufficiently large, the nth prime is 2n log n. b) Show that for n sufficiently large, the nth prime is nlog n)/3. 3) Eplain why the Chebyshev estimates established here imply that there are positive constants C and C 2 such that C log <π)<c 2 for all 2. log

9 More Background Material: Abel s Summation Formula Theorem. For a function a with domain Z +, set A) = k ak). Suppose f is a function with a continuous derivative on the interval [u, v] where <u<v. Then an)fn) =Av)fv) Au)fu) u<n v v u A)f )d. 9 Proof. Clearly, the left-hand side is not changed by replacing u with [u] and v with [v]. One checks that the same is true of the right-hand side. It therefore suffices to consider u and v to be integers. Observe that an) =An) An ). The theorem follows from u<n v an)fn) = v n=u+ An) An ))fn) = Av)fv) Au)fu +) = Av)fv) Au)fu) = Av)fv) Au)fu) v n=u+ v n+ n=u v u An) fn +) fn))an) n A)f ) d. f ) d Homework: ) Let A be a set of positive integers, and let A) = {a A : a }. Suppose that there is a constant c> such that a A a a >clog log for all sufficiently large. Prove that given any t>, there is an t for which A) > c 2 log. Note that I did not say that for all t the last inequality holds.)

10 2) Let n denote an arbitrary positive integer. Recall that ) e =++ 2 2! for all R. 3! a) Using ), show that n! > n/e) n. b) Let = n + in ) and show that the first 2n + 2 terms are each n +) n /n!. Also, show that the remaining terms sum to a number < n +) n /n!. n +)n c) Use b) to show that n! < e n+ 2n + 3). ) n+ n + d) Modify the above to show that n! < 2. e 3) a) Let n be a positive integer. Using Abel summation, prove that n log k = n log n k= n [] d. b) Using a) and the inequality [], prove that n! > n/e) n. c) Do a similar argument to show that n! < nn+ e n. d) Which upper bound on n! is better when n is large: the one given in 2)d) or the one given in 3)c)? Eplain. Derivatives of Sums and Sums of Derivatives Theorem. Let {f n )} be a sequence of differentiable functions such that F ) = f n) converges for > and G) = f n) converges uniformly for >. Then F ) eists and F ) =G)for >. Proof. Let ε>. Let F N ) = N f n) and G N ) = N f n). There is an N = N ε) such that if M N and is any real number >, then Let N and M be N. Then G M ) G) < ε 4. G N t) G M t) < ε 2 for all t>. Fi >. Then there is a δ = δm,), ) such that if h <δ, then F M + h) F M ) G M ) h < ε 4. Fi h <δ. Since F N F M is differentiable on, ), we deduce from the Mean Value Theorem that there is a t between and + h such that F N + h) F M + h)) F N ) F M )) = hf Nt) F M t)) = hg N t) G M t)).

11 We deduce that F N + h) F N ) h F M + h) F M ) h < ε 2. This is true independent of the choice of N N so by letting N tend to infinity, we obtain F + h) F ) F M + h) F M ) h h ε 2. Combining the first, third, and fifth of the above displayed inequalities, we obtain that F + h) F ) G) h <ε. It follows that F ) eists and F ) =G) as desired. Further Elementary Results: Comment on this material. Following classical lines, we show shortly that if lim π)log )/ eists, then it is. The approach below is of some interest in itself, but we note if the goal is simply to establish this information about lim π)log )/, there is an easier way. One can skip the net two sections, and go to Merten s formulas. Then this result on lim π)log )/ follows as a consequence of Merten s formulas and an application of Abel summation see the third homework problem below). The function ζs) and its derivative We use Euler s identity discussed earlier and define the Riemann zeta function to be ζs) = n s = p p s ) for s>. We will also use the von Mangoldt function Λn) which is defined to be log p if n = p k for some prime p and some positive integer k and to be otherwise. In particular, ψ) = Λn). Observe that for s>) n log ζs) = p log p ) s = p m= mp ms. By taking derivatives on both sides of this first equation and recalling the preliminary results above), we deduce that ζ s) ζs) = p p s log p p s = log p p ms p,m = Λn) n s.

12 2 By Abel summation, we now deduce that ζ s) ζs) = s ψ) d s+ for s>. If lim π) log / eists, it is. Theorem. lim inf π) log lim sup π) log. Lemma. lim s +s )ζs) =and lim s +s )2 ζ s) =. Proof of Lemma. For the first limit, use that s = s d < n s < + For the second use an integration by parts to obtain s d = s s. ζ s) = log n n s log = s d + E = s ) 2 + E, where E. Proof of Theorem. Let fs) = ζ s)/ζs). The lemma implies that lim s +s )fs) eists and is. It suffices to show lim inf ψ) ψ) lim sup. Assume the first inequality is incorrect. Then there is a K> and an such that if, then ψ)/>k. Hence, for s>, fs) =s ψ) d>s s+ ψ) K K s+ d+s s d = s ψ) K s+ d+ sk s. Multiplying through by s and letting s +, we obtain a contradiction. In the same manner, one handles the possibility that ψ)/<l< for. Merten s Formulas The following results are due to Merten: n Λn) n = log + O), p log p p = log + O), ψt) t 2 dt = log + O),

13 p p = log log + A + O/ log ), and p ) p B log, where A and B are constants. Discuss what these results mean the notation). Observe that for any positive integer m, 3 e m = j= m j j! > mm m!. For m, we easily deduce that Also, logm!) = p m [log m/ log p] j= m log m logm!) m log m m. [ ] m p j log p = n m where we have used that ψm) = n mλn) =Om)). obtain Λn) = log m + O). n n m [ ] m Λn) = mλn) + Om) n n n m Dividing through by m, we The first of Merten s formulas now follows consider replacing with []). The second formula follows from the first by observing p j=2 log p p j = p log p pp ) is a convergent series. The third formula follows from the first by using Abel summation take an) =Λn) and ft) =/t) and using that ψ) =O) by Chebyshev s estimates). The fourth formula follows from the second by Abel summation take ap) = log p)/p, an) = otherwise, and ft) =/log t). The fifth formula is left as a homework problem. Homework: ) We showed that if { n } is a sequence tending to infinity and lim n ϑ n )/ n or lim n π n ) log n / n eists, then they both do and are equal. Show that this follows from Abel summation you may also use lim π)/ =, but this is not necessary). 2) Prove the formula ) B above. Hints: Use the Maclaurin series for p log p log ) as discussed earlier. You will want to take advantage of another of Merten s formulas.)

14 4 3) Use the fourth formula of Merten above and Abel summation to give an alternative proof of the result π) log π) log lim inf lim sup. Comple Preliminaries: Analytic functions on a region Ω non-empty, open, and connected) The derivative of fz) eists on Ω. All the derivatives of fz) eist on Ω. The function f + iy) =u, y)+iv, y) is such that u and v satisfy the Cauchy- u Riemann equations in Ω: = v u and y y = v. The function fz) may be represented as a power series at each point in Ω. The Identity Theorem If S Ω has an accumulation point in Ω and f and g are analytic functions for which fz) =gz) for all z S, then f g on Ω. Weierstrass Theorem a version of it) Let f s) denote an analytic function in Ω for each. Suppose, as approaches infinity, f s) converges uniformly to fs) on every compact subset of Ω. Then fs) is analytic in Ω and f s) converges uniformly to f s) on every compact subset of Ω. Homework: ) Fi σ R. Let C be a compact set in the region Res) >σ. Prove that there is a σ >σsuch that C is in the region Res) >σ. The Riemann Zeta Function in the Comple Plane: The function ζs) in the right-half plane The definition ζs) = /ns is well-defined for s = σ + it σ and t real) with σ> and defines an analytic function in this region here we interpret n s as e s log n where log refers to the principal branch of the logarithm). It converges uniformly for σ σ >. By Abel summation with an) =andft)=/t s ), we deduce that n s =+ s s {u} du us+ for σ>. {u} By considering Weierstrass Theorem with f s) = du, we deduce that the righthand side is analytic for σ> ecept for a pole with residue at s =. By the Identity us+ Theorem, the right-hand side is the only possible continuation of ζs) to the right-half plane as a meromorphic function). The Riemann zeta function ζs) refers to the right-hand side above when σ>.

15 Euler s identity ζs) = p p s ) holds for σ>. Observe that for σ>, 5 p ) p s + ) p σ ζσ). p It follows that the absolute value of the product in Euler s identity is bounded below by /ζσ). In particular, ζs) is non-zero for σ>. Homework: ) Look at our argument for Euler s identity given for real s>. Eplain how to modify it to show the identity holds for σ> as stated above. The Riemann Hypothesis with ζs) defined as above, ζs) = = σ= /2; discuss some partial results that are known and implications such as on the gap problem for primes) The line σ = Theorem. If t, then ζ + it). Lemma. log z = Relog z). Lemma cos θ + cos2θ) = 2 + cos θ) 2. Proof of Theorem. From Lemma, for σ>, log ζs) = Relog ζs)) = Re p m= ) ) a n mp ms =Re n s = where a n =/m if n = p m for some prime p and a n = otherwise. Hence, log ζ 3 σ)ζ 4 σ + it)ζσ + i2t) = a n cost log n), nσ a n cost log n) + cos2t log n)). nσ The definition of a n and Lemma 2 imply that the right-hand side is. It follows that, for σ>, 4 σ )ζσ) 3 ζσ + it) σ ζσ + i2t) = σ ζ3 σ)ζ 4 σ + it)ζσ + i2t) σ. Assume ζ+it) = with t. We let σ approach from the right above. We have already shown σ )ζσ) approaches also clear from the analytic continuation formulation of ζs)). Observe that ζσ + it)/σ ) approaches ζ + it). Thus, taking a limit of the left-hand side above, we obtain ζ + it) 4 ζ + i2t). On the right-hand side, the limit approaches infinity. Thus, we obtain a contradiction, establishing the theorem.

16 6 Proof of the Prime Number Theorem: The Prime Number Theorem was first established independently by Hadamard and de la Vallée Poussin in 896. It is the following: π) log Theorem. lim =. Homework: ) Let p n denote the nth prime. Using the Prime Number Theorem, prove that lim n p n n log n =. 2) Let ε>. Using the Prime Number Theorem, prove that there is an = ε) such that e ε) < p p<e +ε) for all. 3) Let p j denote the jth prime, and suppose that a,a 2,...,a m are positive integers. If p r is the largest prime factor of a a 2 a m, then we can factor each a k as a k = r j= p e jk) j where each e j k). The least common multiple of the integers a,a 2,...,a m, denoted lcma,a 2,...,a m ), satisfies lcma,a 2,...,a m )= r j= p f j j where f j = ma{e j k) : k m}. Let c denote a constant. Using the Prime Number Theorem, prove that { lcm, 2,..., n) if c> lim n e cn = if c<. For the proof of the Prime Number Theorem, we will make use of previous material together with the following result the Wiener-Ikehara Theorem). Theorem. Let A) be a non-negative, non-decreasing function of, defined for [, ). Suppose that the integral A)e s d converges for σ>to a function fs) which is analytic for σ ecept for a simple pole at s = with residue. Then lim e A) =.

17 The connection. We show here how the Wiener-Ikahara Theorem implies the Prime Number Theorem. Recall that we showed ζ s) ζs) = s ψ) d s+ for s>. For σ >, the right-hand side is analytic by Weierstrass Theorem consider f s) = ψt) s dt). Since ζs) is a non-zero analytic function for σ>, we deduce that the ts+ left-hand side is analytic for σ>. The above equation for real s> now implies by the Identity Theorem that the same equation holds for all comple s = σ + it with σ>. Replacing with e in the integral, we deduce that ) ζ s) sζs) = ψe )e s d for σ>. By our knowledge of ζs) on the line s = + it, the left-hand side of ) is analytic for σ ecept for at s =. We have already seen lim s s )ζs) = and that lim s s )ζ s)/ζs) =. It follows from the Wiener-Ikahara Theorem with A) = ψe ) that lim ψ)/ = which as we have seen implies the Prime Number Theorem. Some preliminary analysis results: Lemma. Let f :[a, b] [, ) C be a continuous function. Suppose that there are positive numbers C and ε such that ft, ) Ce ε for every pair t, ) [a, b] [, ). Then b b ft, ) d dt = ft, ) dt d. a Comment: To prove the lemma, one can consider the double integral of ft, ). This integral is finite, and this justifies such an interchange. Lemma 2. Let f s) be analytic on σ = Res) hence, in an open region containing such s). Let a and b be real numbers with a<b. Then as w +, f + w + iy) converges uniformly to f + iy) for y [a, b]. Proof. Write f + iy) =u, y)+iv, y). Since f is anaytic on σ, each of u and v has continuous partial derivatives for, y) with. Let M be a bound on u, y) and v, y) in the set [, 2] [a, b]. Then with z =+iy and w, the Mean Value Theorem gives fz + w) fz) u + w, y) u,y) + v + w, y) v,y) 2wM. The lemma follows. Lemma 3. Let a and b be real numbers with a<b.letht, w) :[a, b] [, ) C be a continuous function such that ht, w) converges to ht, ) uniformly for t [a, b]. Then b b lim ht, w) dt = ht, ) dt. w + a a a 7

18 8 Proof. Fi ε>. Choose δ>such that if w<δand t [a, b], then ht, w) ht, ) <ε/b a). Then b b b ht, w) dt ht, ) dt ht, w) ht, ) dt<ε. a The result follows. Lemma 4. Let f : R R be a non-negative function for which lim w + f)e w d = a f) d. a t f) d eists. Then Proof. Let ε>. Fi t> such that f) d<ε/4. Let I = f) d. Fi δ> t such that if w<δ, then ε 2I + <e wt. Then f)e w d f) d ε t f)d +2 f)d<ε. 2I + t The result follows. Lemma 5. Let f : R R be a non-negative function for which lim eists. Then Homework: lim w + f)e w d = ) Prove Lemma 5. Do not assume f) d. f) d eists.) w + f)e w d We will use the following weak version of the Riemann-Lebesgue Lemma. Lemma 6. Let a and b be real numbers with a<b.letf:[a, b] C be such that: i) f eists everywhere in [a, b], and ii) b a f t) dt is finite. b Then lim ft)e iyt dt =. y a Comment: Observe that ii) holds if f is continuous on [a, b]. This will be the case for our use of the lemma. Proof. Integration by parts gives b a ft)e iyt dt = ft) eiyt iy b a b f t)e iyt dt. iy a

19 Thus, The result follows. b a ft)e iyt dt 2 fa) + fb) ) y + y b a f t) dt. We will also make use of sin 2 Lemma 7. 2 d = π. There is no real need to discuss the proof as the value of the integral will not come into play; only its eistence will which is clear). A Proof of the Wiener-Ikehara Theorem Set B) =A)e. It follows that 9 fs) s = B) )e s ) d for σ>. Define gs) to be the left-hand side. By the conditions on f, we deduce that g is analytic for σ. Fi ε>, and let ht) =g + ε + it). Fi λ>. Then 2λ 2λ ht) t ) 2λ e iyt dt = t ) )e iyt B) )e ε+it) d dt. 2λ 2λ 2λ We justify that for σ> we can interchange the order of integration by using Lemma. We consider s =+ε/2) and >. Observe that fs) = Au)e us du A) e us du = A)e s. s Thus, B) =A)e sfs)e ε/2). It follows that t ) e iyt B) )e ε+it) Ce ε/2), 2λ where C = sfs) +. Lemma now justifies the interchange of the order of integration. A direct calculation gives 2λ t ) e iy )t dt = sin2 λy )) 2 2λ 2λ λy ) 2. We deduce that ) 2λ ht) t ) e iyt dt = 2 2λ 2λ B) )e ε sin2 λy )) λy ) 2 d.

20 2 Since gs) is analytic for σ, we deduce from Lemma 2 that ht) see its definition) converges uniformly to g + it) for t [ 2λ, 2λ] asεapproaches from the right. It follows from Lemma 3 that lim ε + 2λ 2λ ht) t 2λ ) 2λ e iyt dt = g + it) t ) e iyt dt. 2λ 2λ By Lemma 4, lim e ε sin2 λy )) sin 2 λy )) λy sin 2 v ε + λy ) 2 d = λy ) 2 d = v 2 dv. Since B) is non-negative, Lemma 5 and ) now give lim B)e ε sin2 λy )) ε + λy ) 2 d = B) sin2 λy )) λy ) 2 d λy = B y v ) sin 2 v λ v 2 dv. We let ε + in ). Net, we let y tend to infinity. Observe that 2λ 2λ g+it) t ) e iyt dt = g+it) + t ) 2λ e iyt dt+ g+it) t ) e iyt dt 2λ 2λ 2λ 2λ and we can apply Lemma 6 to each of the integrals on the right. From this and Lemma 7, we obtain λy ) lim B y v ) sin 2 v y λ v 2 dv = π. Eplain why intuitively ) implies we are through. Let a be such that <a<λy. Since A) =B)e is non-decreasing, we get for a v a that e y a/λ) B y a ) e y v/λ) B y v ) e y+a/λ) B y + a ) λ λ λ so that B y a ) e 2a/λ B y v ) B y + a ) e 2a/λ. λ λ λ We now deduce from ) and the fact that B) is non-negative that e 2a/λ lim sup By) ) a sin 2 v y a v 2 dv = lim sup B y a a )e 2a/λ y λ lim sup y a a B y v λ a sin 2 v v 2 ) sin 2 v v 2 dv π. dv

21 The above holds for any positive a and λ. Letting a = λ and letting λ tend to infinity gives that lim sup B). To finish the proof, we show lim inf B). Observe that lim sup B) implies B) c for some constant c and all. We take a = λ again and use that π = lim inf y a c a c λy B y v λ sin 2 v v 2 dv + sin 2 v v 2 dv + ) sin 2 v dv a a v 2 sin 2 v v 2 sin 2 v v 2 ) a dv + lim inf y The theorem follows now by letting λ tend to infinity. a B y v ) sin 2 v λ v 2 dv ) dv + e 2a/λ a lim inf By)) sin 2 v y a v 2 dv. Intermission: Question: Are there infinitely many primes whose decimal representation contains the digit 9? Answer: Yes. Theorem. Given any block of digits d d 2...d n base b 2, there eist infinitely many primes whose base b representation contains the block of digits. We will specialize our arguments to the original question concerning the digit 9. We give three proofs that infinitely many such primes eist. One proof will use the main result of what is to come, one will use the main result of where we have been, and the other won t use much. Each of the arguments easily generalizes to give the above theorem. Before continuing, we note that it is unknown whether or not there eist infinitely many primes whose decimal representation does not contain the digit 9. Actually, it is known; but no proof eists.) What is to come. The net main goal for the course is to establish Dirichlet s theorem that if a and b are relatively prime integers with a>, then there are infinitely many primes of the form an + b. Observe that with a = and b = 9, we can deduce that there are infinitely many primes whose decimal representation contains in fact, ends with) 9. Note that the more general theorem stated above can be done the same way by considering b = d d 2...d n +.) Where we have been. We could instead use the Prime Number Theorem as follows. Lemma. Let ɛ>. Then there is an = ε) such that if, then the interval, + ε] contains a prime. Before proving the lemma, observe that it implies what we want by taking ε =/9 and =9 k. In fact, a similar argument shows there are infinitely many primes whose decimal representation begins with any given block of digits.) Proof. We may suppose that ε and do so. From the Prime Number Theorem, there is an = ε) such that if, then ε ) log <π)< + ε ) log. 2

22 22 Thus, implies π+ε) π) ε ) + ε log2) + ε ) log ε log2) log 2 log )log2)) ε 5 2 log where the last term on the right is a bound on the contribution from the parts involving ε/). We deduce that if is sufficiently large, the interval, + ε] contains a prime. An elementary argument. Let n,n 2,... be the positive integers in increasing order whose decimal representations do not contain the digit 9. Then for N n k < N n k = N N n k j n k < j j= j= 9 j < 9. j It follows that the partial sums of k= /n k form a bounded increasing sequence, and hence the series converges. Since the sum of the reciprocals of the primes diverges, it follows that there are infinitely many primes not among the numbers n k. Hence, there eist infinitely many primes whose decimal representation contains the digit 9. Homework: ) An open problem of Erdős is to show that if {a j } j= is an arbitrary infinite sequence of integers such that i) a <a 2 <a 3 < and ii) j= /a j diverges, then there eist arbitrarily long arithmetic progressions among the a j. More precisely, given a positive integer N, one can find N distinct a j in an arithmetic progression. If true, this would imply there are arbitrarily long arithmetic progressions among the primes, something which as yet is unknown. One approach to resolving the problem might be to consider a sequence {a j } j= satisfying i) and having no N term arithmetic progression and to show that j= /a j must then converge. If one can show this is true regardless of the value of N, then the problem of Erdős would be resolved. The case N = and N = 2 are not interesting. Deal with the following special case with N = 3. Begin with a = and a 2 = 2. Let a 3 be as small as possible so that no 3 term arithmetic progression occurs among the a j then selected. We get a 3 = 4. Choose a 4 now as small as possible avoiding again 3 term arithmetic progressions. Then a 4 = 5. Continue in this way. The net few a j s are,, 3, and 4. Prove that j= /a j converges. Hint: Think base 3.) Algebraic Preliminaries: A group is a set G of objects together with a binary operation satisfying: i) If a and b are in G, then so is a b. ii) If a, b, and c are in G, then a b) c = a b c). iii) There is an element e of G such that a e = e a = a for every a G. iv) For every a G, there is a b G such that a b = b a = e. An abelian group is a group G such that a b = b a for all a and b in G. A group is finite if G is finite. We will simply use ab to denote a b and refer to the binary operation as multiplication unless noted otherwise).

23 Eamples. The set of integers under addtion, the set of non-zero rational numbers under multiplication, and the reduced residue system modulo n under multiplication are all abelian groups. Another eample, is given by {ζ j : j n }where ζ = e 2πi/n and the binary operation is multiplication. Yet another eample is given by {φ,φ 2,φ 3 } under composition where the φ j are defined multiplicatively on the elements of {ζ,ζ 2,ζ 4 } where ζ = e 2πi/7 with φ ζ) =ζ,φ 2 ζ)=ζ 2, and φ 3 ζ) =ζ 4. Finally, consider the multiplicative mappings from the reduced residue system modulo 7 to {ζ j : j 5}where ζ = e 2πi/6 ; show that these mappings form a finite abelian group under multiplication. A simple theorem on finite groups. We will use the fact that if a is an element of a finite abelian group G, then a G = e where e is the identity element in G. Give a proof. Note that the same is true for any finite group. A theorem on finite abelian groups Theorem. Let H be a subgroup of a finite abelian group G. Let a G, and let k be the minimal positive integer for which a k H. Let H = {a j b : b H, j<k}. Then H is a subgroup of G and H = k H. Proof. Let a i b and a j b be elements of H with i, j < k and b and b in H. If i+j<k, then bb H implies a i a j bb = a i+j bb H.Ifi+j k, then i+j k <k. In this case, a k bb H implies a i a j bb = a i+j k a k bb H. Thus, H is closed under multiplication. One also checks that the inverse of a i b is a k i a k b ) H note that if i =, then a i b H H ). The other group properties of H are easily determined to hold, and thus H is a subgroup of G. To prove H = k H, it suffices to show that if a i b = a j b, then i = j. Assume otherwise; we suppose as we may that i>j. Then a i j = b b H contradicts the minimality condition on k. The contradiction completes the proof. 23 Characters: The definition. A character χ of a finite abelian group G is a multiplicative comple valued function, not identically zero, defined on the group. Thus, if a and b are elements of the group, then χab) = χa)χb). Properties of characters: i) If e is the identity element of G and χ any character, then χe) =. This follows from two observations. First, χe) ; otherwise, χa) =χa)χe) = for all a G contradicting the requirement in the definition that χ is not identically zero. Net, χe) = χe)χe), and we can deduce from our first observation that χe) =. ii) If G = n, then χa) isannth root of unity for every a G. This follows from χa) n = χa G )=χe)=. iii) For every a G, χa). See ii).)

24 24 iv) If G = n, then there are eactly n distinct characters defined on G. Let H = {e} where e is the identity element of G. Thus, H is a subgroup of G. If G H, we apply the theorem of the previous section on groups) to construct a new subgroup H 2 of G. Note that H 2 will be a cyclic subgroup of G generated by an element a different from e in G. How a e is chosen doesn t matter. If G H 2, then we again apply the theorem to obtain a new subgroup H 3 of G. We continue defining for i, as long as G H i, the subgroup H i+ = {a j i b : b H i, j<k i }where a i H i and k i is the minimal positive integer for which a k i i H i. We show by induction on i that the number of distinct characters on H i is H i. The result will then follow. We deduce from i) that there is precisely one character defined on H = {e}. Suppose there are precisely H i different characters defined on H i for some i. If G = H i, then we are done as there are no more subgroups to consider). Otherwise, we wish to show that there are precisely H i+ = k i H i different characters defined on H i+. Observe that any character on H i+ is necessarily a character when restricted to H i make use of i) or iii)). We finish the argument by showing that each character χ of H i etends to eactly k i different characters on H i+. Fi χ. Observe first that the definition of H i+ and the multiplicativity of χ imply that χ will be defined on H i+ once the value of χa i ) is determined. On the other hand, the value of χa i ) k i = χa k i i ) is known since ak i i H i ). Call this number γ. Then χa i ) must be one of the k i distinct k i th roots of γ. This shows that there are at most k i different etensions of χ to H i+. On the other hand, defining χa i )tobesuchak i th root of γ is easily shown to produce an etension of χ to H i+. This completes the proof. v) If a G and a e, then there is a character χ defined on G such that χa). In our construction of the H i above, we take H 2 = a. The order of a given in the argument there is k. The construction gives a character χ of H 2 such that χa) is an arbitrary k th root of unity. We choose a k th root of unity other than. This character etends to a character of G with the desired property. vi) The characters of G form a finite abelian group under multiplication. If χ and χ 2 are two characters defined on G, then χ = χ χ 2 means χa) =χ a)χ 2 a) for every a in G. One checks directly that χ is a multiplicative, comple valued function which is not identically zero; hence, the product of two characters is itself a character. Associativity and commutitivity follow from the same properties for multiplication in the set of comple numbers. The identity element, called the principal character, is defined by χa) = for every a G; and this is easily checked. One also checks that if χ is a character, then so is χ defined by χ a) =χa)) and that χ is the inverse of χ. From iv), we now deduce that the characters form a finite abelian group. We note that the group G and the group of characters defined on G are isomorphic; we do not need this fact and so do not bother with the details of an eplanation. vii) The following holds: { if χ is not principal χa) = G if χ is principal. a G Observe that in the case that χ is principal, the result is trivial. If χ is not the principal

25 character, then there is a b G such that χb). The proof follows by multiplying the sum, say S, byχb) and using that {ab : a G} = G. Thus, χb)s = S so that S =. viii) The following holds: { if a e χa) = G if a = e. χ If a = e, the result follows from iv). If a e, then by v) there is a character χ such that χa). If the sum above is S, we obtain χa)s = S and the result follows along the same lines as vii). Dirichlet characters. A Dirichlet character χ is a character on the reduced residue system modulo n etended to the complete system modulo n by defining χa) = whenever a and n are not relatively prime. Eamples of Dirichlet characters. Discuss the two Dirichlet characters modulo 6, the four Dirichlet characters modulo 8, and the twelve Dirichlet characters modulo 36 by making Dirichlet character tables in the first two cases and eplaining how to go about doing the same in the latter by defining χ5) to be a sith root of unity and χ35) to be ±). Homework: ) a) Make a Dirichlet character table of the eight characters modulo 5. b) Make a Dirichlet character table of the eight characters modulo 2. Dirichlet Series: The definition. A Dirichlet series is a series of the form a n/n s where a n and s denote comple numbers we interpret n s as e s log n where log refers to the principal branch of the logarithm). Preliminary results. Theorem. Fi θ,π/2). If the series a n/n s converges for s = s C, then it converges uniformly in {s : args s ) θ}. Proof. By considering b n = a n n s so that a n/n s = b n/n s s, we see that it suffices to prove the theorem in the case that s =. Thus, a n converges and its partial sums form a Cauchy sequence. Fi ε>. Consider a positive integer M such that M<n a n <εfor all M. We definie a n = a n if n>m and a n =ifn M. Let N>M. We apply Abel summation with A) = n a n and f) = s to obtain ) M<n N a n n s = AN)fN) AM)fM)+s N M A) d. s+ The definition of A) implies that A) <εfor all M. We write s = σ + it and obtain N A) N d s+ A) N s+ d ε σ+ d = ε σ M σ ) N σ. M M M 25

26 26 Note that From ), we deduce easily that M<n N a n n s < s σ cos θ. 4 s ε σm σ 4 s ε σ 4ε cos θ. Therefore, the partial sums of a n/n s form a Cauchy sequence and hence the series converges. The above inequality in fact implies uniform convergence. Theorem 2. If a n/n s converges for s = s = σ + it, then it converges in the half-plane s = σ + it with σ>σ. Furthermore, in this case, the convergence is uniform on every compact subset of the half-plane with σ>σ. The proof of Theorem 2 is clear given Theorem ). The smallest value of σ is called the abscissa of convergence for the Dirichlet series. Observe that if σ is the abscissa of convergence for a Dirichlet series a n/n s and if σ is finite, then the series converges if σ = Res) >σ and diverges if σ<σ. What happens on the line σ = σ is unclear. It should be noted that σ could be + or with, for eample, a n = n! and a n =/n!, respectively). Theorem 3. Let σ be the abscissa of convergence for the Dirichlet series a n/n s. Then the series represents an analytic function in the half-plane σ>σ and its derivatives can be computed by termwise differentiation. In the comple preliminaries section of the notes, we introduced a theorem of Weierstrass. By considering f s) = n a n/n s and applying Theorem 2, the above result follows. Theorem 4. Let σ R. Suppose a n/n s and b n/n s both converge for σ>σ and that they are equal on some non-empty open set in this half-plane. Then a n = b n for all n. Proof. Let c n = a n b n. The conditions in the theorem, Theorem 3, and the Identity Theorem imply that c n/n s is identically in the half-plane σ>σ. We wish to prove that c n = for all n. Assume otherwise, and let M be minimal with c M. Fi for the moment σ>σ. Observe that c n/n σ converges so that c n /n σ < for n sufficiently large, say n N M + 2. Now, for u>, c n /n σ+u < /n u for n N. We obtain from n=m c n/n σ+u = that c M M σ+u n=m+ N c n n σ+u ) M +) σ+u c n + n=m+ n=n n u C M +) u, for some constant C. Multiplying through by M σ+u and letting u tend to infinity gives a contradiction and establishes the theorem.

27 Part I of the Proof of Dirichlet s Theorem: χn) The Dirichlet series Ls, χ) = where χ denotes a character modulo a n s positive integer m converges for σ > ifχis not the principal character. To see this, observe that property vii) of characters easily implies n χn) is bounded as goes to infinity. An application of Abel summation now shows that Ls, χ) converges for σ>. It is easy to see that Ls, χ) diverges for σ <. It follows that is the abscissa of convergence. In particular, Ls, χ) is analytic in the region σ>. The connection between Ls, χ) and ζs) when χ is the principal character. First, for an arbitrary Dirichlet character χ, a proof similar to that done before gives Ls, χ) = p χp) ) p s for σ>. For the principal character χ modulo m, we obtain Ls, χ )= p m p s ) p ) p s = p ) s ζs). p m 27 It follows that Ls, χ ) is analytic for σ> ecept for a simple pole at s = with residue /p)). p m For every Dirichlet character χ, Ls, χ) for σ>. A proof similar to that given for the analogous result for ζs) can be used here. The logarithm of Ls, χ). For an arbitrary Dirichlet character χ modulo m, we define wls, χ)) = p k= χp k ) kp ks for σ>. The right-hand side is what we would obtain from using the Maclaurin series for log ) with the product formulation of Ls, χ) if we incorrectly) assume logz z 2 ) = log z +log z 2 for comple numbers z and z 2. Locally, w behaves like the logarithm of Ls, χ) but globally it may not correspond to any branch of the logarithm. In particular, since for the principal branch of the logarithm, log z) = z z 2 /2 z 3 /3 for z <, it follows that ep z z 2 /2 z 3 /3 )= zfor z < so that ep p k= χp k ) ) kp ks = χp) ) p s for σ>. p Thus, ep wls, χ)) ) = Ls, χ) for σ>.

28 28 Observe that w is defined as a Dirichlet series which is analytic for σ >. In particular, it is differentiable, and we have by taking derivatives above that d ds Ls, χ)) = L s, χ) Ls, χ) for σ>. Hence, for fied s = σ> and c>σ, we obtain ) wls, χ)) = c s L u, χ) Lu, χ) du + log Lc, χ). Homework: ) a) Let χ be a character on a group G. Let χ be the function defined by χg) =χg) the comple conjugate of χg)). Prove that χ is a character on the group G. b) Let Ls, χ) be the Dirichlet L-series corresponding to a Dirichlet character χ modulo a positive integer m. Let χ denote the principal character modulo m. Recall that Ls, χ) is analytic for σ>ifχ χ and that Ls, χ ) is analytic for σ> ecept for a simple pole at s =. Suppose L,χ), χ where the product is over all Dirichlet characters χ modulo m. Also, suppose χ is a character for which χa) ± for some integer a. Eplain why L,χ). Part II of the Proof of Dirichlet s Theorem: We will show in the net section that L,χ) ifχis a non-principal Dirichlet character. We use this fact for the remainder of this section to show how Dirichlet s theorem on primes in an arithmetic progression follows. Observe that from ) above, for χ a non-principal character, we deduce from the fact that Ls, χ) and L s, χ) are analytic for σ> and L,χ) that wls, χ)) remains bounded as s +. The proof of Dirichlet s Theorem assuming L,χ) ifχis a non-principal Dirichlet character. Let a and m be integers with m> the case m = is trivial) and gcda, m) =. We prove there are infinitely many primes p a mod m) by showing that the sum of the reciprocals of such primes diverges. Let b Z with ab mod m). Observe that b mod m) if and only if a mod m). Consider the Dirichlet characters modulo m, and let r denote the number of them. By property viii) of characters, it follows that { r if p a mod m) χpb) = otherwise. We consider s = σ> and define E by χ ) wls, χ)) = p χp) p s + E

29 29 so that E = p k=2 χp k ) kp ks. A simple estimate gives that E. absolute value by r that From ), we obtain for some E bounded in χb)wls, χ)) = χ χ χb) p χp) p s + E = p χ χpb) p s + E = r p a mod m) p s + E. We now let s + and observe that each term in the sum on the left remains bounded ecept for the summand involving the principal character and this summand increases in absolute value to infinity. This implies that the right-hand side must increase in absolute value to infinity so that the sum of the reciprocals of the primes p a mod m) diverges. Part III of the Proof of Dirichlet s Theorem: We begin with some preliminaries: Lemma. Suppose fz) is analytic in a region Ω containing a point z. Suppose the disk D = {z : z z <r}where r>is in Ω. Then the series fz )+ f z )! converges in D to fz). z z )+ f z ) 2! z z ) 2 + f z ) z z ) 3 + 3! The lemma asserts that fz) is equal to its Taylor series representation about z in any open disk centered at z that is contained in Ω. We omit the proof of this lemma but use it to establish the following result of Landau discuss the connection with ζs) as an eample to help clarify the theorem). Theorem. Let fs) be analytic for σ = Res) >. Suppose that there is a Dirichlet series a n /n s with abscissa of convergence σ that equals fs) for σ>ma{,σ }. If a n for every n, then σ. Proof. Assume σ >. Since fs) is analytic for σ>, fs) can be represented as a Taylor series about σ + that by Lemma ) converges in a disk of radius > + 2 σ centered at σ +. For k a positive integer, the Dirichlet series representation of fs) gives us f k) a n log n) k σ +)=. n +σ

30 3 Thus, inside this disk, fs) = = = k= k= f k) σ +) s σ ) k k! s σ ) k a n log n) k k! n +σ + σ s) k k= k! a n log n) k n +σ. Fi s = u σ /2,σ ) in the disk). Then the double sum converges and, since the terms are non-negative, it converges absolutely. We may therefore rearrange the order of the summations to obtain that fu) = a n n +σ k= + σ u) k log n) k k! = a n n +σ e+σ u)log n) = This is impossible as u is to the left of the abscissa of convergence for a n/n s. The theorem follows. Lemma 2. If a Dirichlet series a n /n s converges for s = s, then the series converges absolutely for Res) > Res )+. Proof. Let σ = Res ), and consider s = σ+it with σ>σ +. Convergence at s implies that lim n a n /n σ =. Thus, a n /n σ < for n sufficiently large. Since σ>σ +, it follows by comparison with /nσ σ that a n/n s = a n/n σ n s σ ) converges absolutely. The product of two Dirichlet series a n/n s and b n/n s is defined to be the Dirichlet series c n/n s where c n = kl=n a kb l here, k and l represent positive integers). If the Dirichlet series involving a n and b n both converge for s = σ + it with σ > σ, then they converge absolutely for σ > σ + by Lemma 2. It follows that the Dirichlet series representing their product will converge absolutely for σ>σ +. The product will, therefore, converge in this same region we view the product then as converging in a possibly smaller region then the initial two Dirichlet series). Observe also that if we consider a n/n s) k for k any positive integer and epanding this product in the obvious way), it is easy to see that independent of k the resulting series converges in fact, absolutely) for σ>σ +. Note that the values of a series representing the product of two Dirichlet series is in fact equal to the product of the values of the two Dirichlet series if the series involved converge absolutely. a n n u. Homework: ) Prove that ζs) 2 = integer divisors of n. dn) n s for σ> where dn) denotes the number of positive

31 2) Recall that we showed ζ s) ζs) = Λn) n s for s>. Eplain why this implies that d n Λd) = log n. Hint: Write down the series representation for ζ s).) 3) Prove that ) n n s =2 s )ζs) for s = σ + it with σ> and s ). Hint: It may help to consider the case σ> first, but don t forget to address the case σ>.) We are now ready to complete the proof of Dirichlet s Theorem. As we saw before, we need only justify the following. Theorem 2. If χ is not the principal character modulo an integer m>, then L,χ). Proof. Recalling the definition of wls, χ)) from the previous section Part II of the Proof of Dirichlet s Theorem), we define for σ>, 3 Qs) = χ wls, χ)) = χ p k= χp k ) kp ks, where the sum on χ is over all characters modulo m. The right-hand side converges absolutely for σ>so that we can rearrange the order of summation. Let r denote the number of characters modulo m. Then by making the inner sum be over χ and using property viii) of Dirichlet characters, we deduce that Qs) = p k< p k mod m) r kp ks = a n n s, where a n = r/k if n = p k mod m) k Z + ) and a n = otherwise. This last series converges absolutely for σ>. Let d be such that u d mod m) for every integer u with gcdu, m) = so we may take d = r = φm), but we needn t be this precise). Then a n = r/d whenever n = p d and p m. It follows that the Dirichlet series a n/n s does not converge for s =/d use that the sum of the reciprocals of the primes diverges and that each a n ). Hence the abscissa of convergence for a n/n s is in [/d, ]. Now, consider fs) =e Qs) =+Qs)+ Qs)2 2! + Qs)3 3! +. The powers of Qs) are themselves Dirichlet series, and each power converges absolutely for σ > since each a n. Also, a = implies that any given /n s appears as a