INTRODUCTORY LECTURES COURSE NOTES, One method, which in practice is quite effective is due to Abel. We start by taking S(x) = a n

Transcription

1 INTRODUCTORY LECTURES COURSE NOTES, 205 STEVE LESTER AND ZEÉV RUDNICK. Partial summation Often we will evaluate sums of the form a n fn) a n C f : Z C. One method, which in ractice is quite effective is due to Ael. We start y taking Sx) = and oserving that n x a n Sn) Sn ) = a n. Using this we see that for integers B > A a n fn) = fn)sn) Sn )) = fn) A <n B =fb)sb) fa)sa) fn + )Sn) A <n B Sn)fn + ) fn)). For an integer n and n x < n + one has Sx) = Sn). So if f is continuously differentiale we can use the fundamental theorem of calculus to see that A <n B B Sn)fn + ) fn)) = = = Sn) n=a n B n+ n=a B A n n+ f x) dx Sx)f x) dx Sx)f x) dx. This imlies the following formula for artial summation Date: March 28, 205.

2 2 STEVE LESTER AND ZEÉV RUDNICK Theorem. Partial summation). Suose that f : R C is continuously differentiale. Then a n fn) = fb)sb) fa)sa) B A Sx)f x) dx. Remark. There is some sutlety with endoints here. Notice that slightly altering the values of A and B may leave the left-hand side of the formula unchanged. As a consistency check verify that the value of the right-hand would also e unaltered. Examle. Evaluate n N log n. Take a n =, fn) = log n, Sx) = x. Here and throughout x is the floor function and equals the largest integer x. The artial summation formula gives N x log n = N log N x dx n N =N log N N + Olog N). For a comlex variale s the Riemann zeta-function ζs) is given y ζs) = n= n s Res) > ). Riemann oserved that the analytic roerties of ζs) are closely related to the distriution of the rime numers and amongst other things) showed that ζs) has an analytic continuation to C \ {}. We will rove Theorem.2. The Riemann zeta-function admits an analytic continuation to the half-lane Res) > 0 excet for a simle ole at s =. Morever for Res) > 0 one has ζs) = s s s {x} dx xs+ where {x} = x x. Proof. Let s e a comlex variale. Using artial summation with Sx) = x and fx) = /x s we get that n N n s = N N N s + s x dx. xs+

3 Take N to get that for Res) > INTRODUCTORY LECTURES 3 x ζs) =s x = s s s dx s+ {x} dx. xs+ Note that the right-hand side is analytic in the half-lane Res) > 0 excet for a simle ole at s =. This rovides the analytic continuation of ζs) to Res) > Cheyshev s theorem and Merten s formulas The rime numer theorem states that πx) = rime = x log x + O = Lix) + O ) x log x) 2. x ex ) log x) For our uroses the weaker estimate of Cheyshev will often e sufficient. Theorem 2. Cheyshev s Theorem). There exist constants c < < C such that cx Cx πx) log x log x. Remark. Let ψx) = n x Λn). Using artial summation Cheyshev s estimate is equivalent to c x ψx) C x. We will rove Cheyshev s Theorem in this form. Proof. Recalling Λ = log we have log n = Λa) = Λa) n x n x a=n x a x/ = x ) x ) ψ = ψ. x Therefore ) x ) ψ = x log x x + Olog x).

4 4 STEVE LESTER AND ZEÉV RUDNICK Aly ) twice to get that 2) ψ ) 2N 2 ψ ) N =2N log 2N 2N 2N log N N) + Olog N) =N log 4 + Olog X). Comining the even terms from the first sum with the second sum gives ) )) 2N N 3) ψ ψ = N log 4 + Olog N). 2 The function ψx) is non-decreasing so each term in the aove sum is ositive. Thus droing all ut the first term 4) ψ 2N) ψ N) N log 4 + Olog N). Using this relation at N = x/2, x/4, x/8,..., x/2 A where A = log x/ log 2 and summing gives A x ) x )) ψ 2 ψ 2 x log Olog x)2 ). Therefore 5) ψ x) x log 4 + Olog x) 2 ). Next rewrite 3) to see ψ2n) ψ N ) )) 2N ψ = N log 4 + Olog N). 2 + Every term in the sum on the right hand side is ositive so that alying this at N = x/2 ψx) x log 2 + Olog N). From the roof it follows that y 3) and 5) Therefore, ψ2x) ψx) x log 4 ψ2x/3) + Olog x) ) 3 log 4 x + Olog x) x< 2x log 2x x< 2x log ) ψ2x) ψx) + O x log x log 2x ) x 3 log 4 + o)). log x

5 INTRODUCTORY LECTURES 5 Corollary 2.2 Bertrand s ostulate). For each real numer x there is a rime numer in the interval [x, 2x]. Remark. Bertrand s ostulate has een significantly imroved. For any sufficiently large x it is known that there exists θ < such that there is a rime numer in every interval of the form [x, x+x θ ]. The est known result in this direction gives θ = 2/40 and it is conjectured that this should hold for any θ > 0. Using the rime numer theorem and artial summation it is straightforward to check that x = dt + o)) = log log x + o)). t log t 2 However, in this instance Cheyshev s theorem suffices to estalish Theorem 2.3 Mertens formulas). We have a) = log log x + O). ) log = log x + O). c) ) log x. Remarks. For f, g > 0 the notation fx) gx) means there exist constants c, c 2 such that c gx) fx) c 2 gx) for all x under consideration. From art c) it immediately follows that φn) n/ log log n, n 3). To see this note that since the numer of rime divisors of n is C log n for some C > ) we have φn) n = n ) C log n ) log log n. Additionally, it is ossile to give more recise formulas than those given aove. In articular, it is known that = log log x + + O/ log x) where is a certain asolute constant. Also, ) e γ log x, where γ is Euler s constant. Proof. We first will estalish ). The argument is similar to the one given to rove Cheyshev s theorem. Use the relation log = Λ and switch order

6 6 STEVE LESTER AND ZEÉV RUDNICK of summation x log n = Λa) x n x n x a=n = Λa) x a x x/a = Λa) ψx) + O a x a x ). Evaluate the left-hand side using artial summation and aly Cheyshev s theorem to get Λa) = log x + O). a Oserve that a x a x Λa) a = log + n x n 2 log. The second sum is clearly O). This gives ). Once again ounding the higher rime owers we see n x = n x Λn) n log n + O). Now use artial summation with a n = Λn)/n, and fx) = /log x) to get Λn) n log n = x log t + O)) log x + O)) + log x tlog t) 2 dt = log log x + O). To estalish art c) we note that ) = ex ) log = ex + O) = ex log log x + O)) log x. 2