INTRODUCTION TO REAL ANALYSIS II MATH 4332 BLECHER NOTES

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1 INTRODUCTION TO REAL ANALYSIS II MATH 4332 BLECHER NOTES You will be expected to reread and digest these typed notes after class, line by line, trying to follow why the line is true, for example how it follows from previous lines. I suggest you add a check mark after you have read and understood the line, add extra explanation or pictures to yourself if needed. Add a question mark next to any line you cannot follow, and ask me or the TA about it. That is why I have given wide margins on every page. Also memorize definitions as you read. The best advice I can give to ensure success in this class is to do this reading properly. In my experience, the class becomes much much more difficult if you do not do it. This kind of detailed reading is not without pain, but it will help reconfigure your brain to internalize the kind of logic and proofs that are needed in this subject (and in other math proof courses). The way I will monitor if you are doing all this, and this is the first semester I ve tried this, is to collect your notes occasionally to see how much you have scribbled on them as above. There will also be an easy-quiz from time to time, named as such because it will be easy for anyone who has been reading the notes as suggested. The problems marked * do not have to be turned in by the undergraduates in the class. Chapter 0. Series of numbers In this course/these notes we will write N for the natural numbers N = {, 2, 3, }, and N 0 for the whole numbers N 0 = {0,, 2, 3, }, and Z for the integers Z = {, 2,, 0,, 2, 3, }. We usually reserve the symbols n, m for natural numbers or integers. The rational numbers are Q = {m/n : m, n Z, n 0}. The real numbers are written as R, and R + = {x R : x 0}. The notation A B or A B means that A is a subset of set B, (that is, x A x B). Of course A = B if A B and B A (that is, x A x B). Complement: A c = {x : x / A}. We say that sets A and B are disjoint if A B = (no common elements). Functions: f : A B means that f is a function from domain A into the codomain B. Image: if C A, and f : A B, then f(c) = {f(x) : x C}. This is a subset of the codomain of f, called the image of C under f. Pre-image: If f : A B, and D B, then f (D) = {x A : f(x) D}. This is called the pre-image of D under f. We say that f : A B is surjective, or onto if f(a) = B.

2 2 INTRODUCTION TO REAL ANALYSIS II MATH 4332 BLECHER NOTES We say that f : A B is injective, or one-to-one if f(x ) = f(x 2 ) x = x 2. We say that f : A B is bijective if it is one-to-one and onto. In this case there is an inverse function f : B A with f (y) = x iff f(x) = y, for x A, y B. See wikipedia for more detail.. Limsup and liminf Definition. The limit superior of a sequence (s n ), is the number limsup n s n = lim n {sup{s k : k n}}. Sometimes this is written as lim n s n. The limit inferior is liminf n s n = lim n {inf{s k : k n}}. Sometimes this is written as lim n s n. Ex. Find the limsup and liminf of (s n ) where s n = ( ) n + n. Solution. Make a table of the terms in the sequence: 0, 3 2, 2 3, 5 4, 4 5, 7 6, 6 7, 9 8,. Procedure: explained in class to get the new row in the table 3 2, 3 2, 5 4, 5 4, 7 6, 7 6, 9 8,. (Each entry in this row is the supremum of the numbers above and to the right of it in the original row.) Note that this new row is a decreasing sequence. Every decreasing sequence has a limit (which is possibly ± ). In our example, this limit is. This limit is exactly limsup n s n. So in our example, limsup n s n =. Similarly to find the liminf, make a third row of the table,,,,,. (Each entry in this row is the infimum of the numbers above and to the right of it in the original row.) Note that this third row is always an increasing sequence. Every increasing sequence has a limit (which is possibly ± ). In our example, this limit is. This limit is exactly liminf n s n. So in our example, liminf n s n =. Ex. Find the limsup and liminf of (s n ) where s n = 6n+4 7n 3. Solution. Do this as an exercise. What the limsup and liminf are good for:

3 INTRODUCTION TO REAL ANALYSIS II MATH 4332 BLECHER NOTES 3 First, they always exist, unlike the limit. For example, in the first example above, lim n s n does not exist. But we were able to compute the limsup and liminf. They always exist because as we saw in the example, they are limits of monotone sequences, which we know always exist. They behave similarly to the limit. That is, they obey laws analogous to the rules we saw in Calculus II and 3333 for limits. We will write down some of these laws momentarily. They can be used to check if the limit exists. In fact lim n s n exists iff liminf n s n = limsup n s n. So if liminf n s n limsup n s n then we may conclude that lim n s n does not exist. Recall that in Calculus II there were certain tests which involve the limit of a sequence, such as the ratio and root test, limit comparison test, and the fundamental fact about power series. We shall see that one can improve these tests by using the limsup and liminf instead of the limit. For example, the fundamental fact about power series states that a power series k=0 c kx k converges absolutely for all points in a certain interval ( R, R), and it diverges whenever x < R or x > R. The number R is called the radius of convergence, and in Calculus II you are given the for- mula R = lim n n c n. But you were not told in Calculus II what to do if this limit does not exist. In fact one should use the formula R = limsup n n c n. This always exists, as we remarked above, and now the test always works. Theorem.. Let (s n ) and (t n ) be sequences of real numbers. () liminf n s n limsup n s n ; and lim n s n exists iff liminf n s n = limsup n s n. (2) liminf n ( s n ) = limsup n s n. (3) If s n t n for all n, then limsup n s n limsup n t n, and liminf n s n liminf n t n. (4) limsup n (Ks n ) = K limsup n s n, and liminf n (Ks n ) = K liminf n s n, if K 0. (5) limsup n (s n +t n ) limsup n s n +limsup n t n, and liminf n (s n +t n ) liminf n s n + liminf n t n. These inequalities are equalities if (t n ) converges. (6) If a = limsup n s n is finite, and if ɛ > 0, then there exists an N such that s n < a + ɛ for all n N. Also for every N there exists a k > N with s k > a ɛ. That is, there are infinitely many terms in the sequence which are greater than a ɛ. (We remark that these properties actually characterize limsup n s n ; and variants of these statements are easily formulated for liminfs.) (7) If a = limsup n s n then a subsequence of (s n ) converges to a. Similarly for liminf.

4 4 INTRODUCTION TO REAL ANALYSIS II MATH 4332 BLECHER NOTES Proof. () Clearly inf{s k : k n} sup{s k : k n}, and so lim n inf{s k : k n} lim n sup{s k : k n}. We also see from this that if these numbers are finite then liminf n s n = limsup n s n iff 0 = limsup n s n liminf n s n = lim n (sup{s k : k n} inf{s k : k n}). By the definition of limit, this happens iff given ɛ > 0 there exists an N such that But the latter is equivalent to sup{s k : k n} inf{s k : k n} ɛ, n N. s m s n ɛ, m, n N. This is saying (s n ) is Cauchy, which is equivalent to saying that lim n s n exists. If liminf n s n = limsup n s n =, then since inf{s k : k n} s n we see lim n s n =. Similarly for the case. (2) From 3333 we have inf{ s k : k n} = sup{s k : k n}. Taking the limit of these as n we get liminf n ( s n ) = limsup n s n. (3) sup{s k : k n} sup{t k : k n}. Taking the limit of these as n we get limsup n s n limsup n t n. (4) From 3333 we have sup{ks k : k n} = K sup{s k : k n}. Taking the limit of these as n we get limsup n (Ks n ) = K limsup n s n. (5) From 3333 we have sup{s k +t k : k n} sup{s k : k n}+sup{t k : k n}. Taking the limit of these as n we get limsup n (s n + t n ) limsup n s n + limsup n t n. Similarly for the liminf case. If (t n ) converges to t R then given ɛ > 0 there exists an N such that t n > t ɛ for all n N. Then sup{s k + t k : k n} sup{s k + t ɛ : k n} = sup{s k : k n} + t ɛ, i for n N. Taking the limit of these as n we get limsup n (s n + t n ) limsup n s n +lim n t n ɛ, for all ɛ > 0. So limsup n (s n +t n ) = limsup n s n +limsup n t n. (6) If a = limsup n s n = lim n sup{s k : k n} is finite then from 3333 given ɛ > 0 there exists an N such that a ɛ < sup{s k : k n} < a+ɛ for all n N. So s n sup{s k : k n} < a + ɛ for all n N. This proves the first assertion. Similarly, sup{s k : k n} > a ɛ for all n N, which means that there must exists a k > n with s k > a ɛ. (7) If we take ɛ = n in the argument in (6), we see that given any m N we have a n < sup{s k : k m} < a + n. So given any m N there exists a k m with s k a < n. Applying this with n =, m = N, choose n with s n a <. Applying this with n = 2 and m > n, choose n 2 > n with s n2 a < 2. Applying this again with n = 3 and m > n 2, choose n 3 > n 2 with s n3 a < 3. Continuing in this way produces a subsequence of (s n ) converging to a. The proofs of (3) (7) above in the liminf cases are similar.

5 INTRODUCTION TO REAL ANALYSIS II MATH 4332 BLECHER NOTES 5 (The last theorem used to be Homework, due Thursday 22 January. grading scale was Q [2 points for first part, 9 for second part], Q2 [3 points], Q4 [2 points], Q6 [6 points for first part, 3 for second part]. Total [25] points for correctness, [5] points for completeness in Q 3, 5, 7.) 2. Infinite series of numbers The following from Calculus II is almost all review, so we will move quickly. You may need to read it carefully several times. You could also look up several of these topics on wikipedia. From Calculus II: An infinite series is an expression of the form a k = a m + a m+ + a m+2 + ( ) k=m Let us call this expression (*). The a k here are real (or complex) numbers. What does expression (*) mean? In fact we shall see shortly that the expression means two things. or Usually m = 0 or, that is, (*) usually is a 0 + a + a 2 + a + a 2 + a 3 +. We call the number a k the kth term in the series. Sometimes we will be sloppy and write k a k when we mean (*). The most important question about an infinite series, just as for an infinite sequence, is ) does the series converge? and 2) if it converges, what is its sum? We will explain these in a minute. In fact an expression like (*) has two meanings: Meaning # : A formal sum. That is, it is a way to indicate that we are thinking about adding up all these numbers in the expression (*), in the order given. It does not mean that these numbers do add up. Before we go to Meaning # 2, let me say how you add up all the numbers in an infinite series. To do this, we define the nth partial sum s n to be the sum of the first n terms in the series. In this way we get a sequence s, s 2, s 3, called the sequence of partial sums. For example, for the series k=0 a k, we have s n = n k=0 a k. We say the original series converges if the sequence {s n } converges. If it does not converge then we say it diverges. We call lim s n the sum of the n series if this limit exists. The

6 6 INTRODUCTION TO REAL ANALYSIS II MATH 4332 BLECHER NOTES Meaning # 2: k a k = lim n s n if this limit exists. Cauchy test: k a k converges iff given ɛ > 0 there exists an N 0 such that a n+ + a n a m < ɛ whenever m > n N. [Proof: Since s m s n = a n+ + a n a m, this is just saying that the partial sums s n = n k= a k are a Cauchy sequence. And we know from Math 3333 that a sequence converges iff it is a Cauchy sequence.] In a sum like k= (k+)k, the k is a dummy index. That is, it is only used internally inside the sum, and we can feel free to change its name, to j= (j+)j, for example, In a series k=m a k let us call m the starting index. Thus for example, the starting index of k=2 k k is 2. Any series can be renumbered so that its 2 starting index is 0. That is, any infinite series may be rewritten as k=0 a k. For example, k=m a k, which is the same as a m + a m+ + a m+2 +, can be relabelled by letting j = k m, or equivalently k = j + m. Then k=m a k = j=0 a j+m. Example: Rewrite k=2 k k 2 as a series k=0 a k. Solution. Letting j = k 2, so that k = j + 2, the sum becomes j=0 j + 2 (j + 2) 2 = j=0 j + (j + 2) 2 Of course j is dummy so we can rewrite this as k=0 k+ (k+2) 2. There is no reason of course why we chose 0 for the starting index. One can make all series begin with the starting index if you wanted to, by a similar trick. However it is convenient to fix one starting index, so it may as well be 0. Many of the following results are therfore phrased in terms of series k=0 a k. Geometric series: This is a series of form c + cx + cx 2 + cx 3 +, or k=0 cxk, for constants c and x. We call x the constant ratio of the geometric series. Note that if you divide any term in the series by the previous term, you get x. We assume c 0, otherwise this is the trivial series with sum 0. The MAIN FACT about geometric series, is that such a series converges if and only if x <, and in that case its sum is c x.

7 INTRODUCTION TO REAL ANALYSIS II MATH 4332 BLECHER NOTES 7 [Proof: If c = 0 then the series is and the result is obvious. So we can assume that c 0. The sum of the first n terms is s n = c + cx + cx cx n = c( + x + x x n ). There is a well known result in algebra which says that ( + x + x x n )( x) = x n (to prove it multiply out the parentheses and cancel). Thus if x then + x + x x n = xn x, so that s n = c + cx + cx cx n = c xn x. This is the important formula for the sum of n terms of a geometric series. The only thing that depends on n on the right hand side here is the x n, which converges to 0 if x <, and diverges otherwise. If x = then s n = c + c + + c (n times) which equals nc. Thus lim n s n = c x if x <. If x then {s n } diverges, so that the original series diverges.] FACT: If k=0 a k and k=0 b k both converge, and if c is a constant, then: k=0 (a k + b k ) converges, with sum k=0 a k + k=0 b k; k=0 (a k b k ) converges, with sum k=0 a k k=0 b k; k=0 (ca k) converges, with sum c k=0 a k. [Proof: We just prove the first and third, the second is quite similar. The nth partial sum of k=0 (a k + b k ) is n k=0 (a k + b k ) = n k=0 a k + n k=0 b k. By a fact about sums of limits of sequences from 3333 or 433, this converges, as n, to k=0 a k + k=0 b k. Similarly the nth partial sum of k=0 (ca k) is n k=0 (ca k) = c n k=0 a k. By 3333, this converges, as n, to c k=0 a k.] For any positive integer m we can write k=0 a k = (a 0 +a + +a m )+ k=m a k. Indeed k=0 a k converges if and only if k=m a k converges. If these series converge, then their sum also obeys the rule: a k = (a 0 + a + + a m ) + a k. k=0 k=m [This is because the nth partial sum of the k=0 a k series, and the nth partial sum of the k=m a k series differ by a fixed constant, namely a 0 + a + + a m.]

8 8 INTRODUCTION TO REAL ANALYSIS II MATH 4332 BLECHER NOTES From the last fact it follows that the first few terms of a series, do not affect whether the series converges or not. It will affect the sum though. The Divergence Test: If lim a k 0 then the series k a k diverges. k A matching statement (the contrapositive): If k a k converges, then lim a k = 0. k [Beware: If lim k a k = 0 we cannot conclude that k a k converges. [Proof: Suppose that k=0 a k = s. If s n is the nth partial sum then s n s as n. Clearly s n+ s too, as n. Thus a n = s n+ s n s s = 0.] Homework 2 (due Tuesday 27 January). () If k a k converges define the tail of the series to be the sequence whose nth term is k=n a k. Prove the tail converges to 0 as n. (2) If y R write [y] = max{n Z : n y}. If x [0, ) write a = [0x], a 2 = [00(x a 0 )], a 3 = [000(x a 0 a2 a n 0 converges, Prove that x = n n= each k and that n= that there is no N such that a k = 9 for all k N. 00 )],. Prove that 0 a k 9 for a n 0. Prove n (3) Continuing with the last question, if x = a n n= 0 where a n n {0,,, 9} then we write x = 0.a a 2 a 3 and call this a decimal expansion of x (or base 0 expansion). Prove that the decimal expansion of x [0, ) is unique provided that we insist that there is no N such that a k = 9 for all k N. (4) Prove that any nonzero real number may be written as ± a n n=n 0 for n some integer N, and with a n {0,,, 9} for all n, and a N 0; and that this representation is unique provided that we keep the recurring 9 convention in question (2). [Remark: Questions 2 4 can be found in many places on the internet under decimal expansion ; or see e.g. the Appendix B to Tao s Analysis I. It can also be done for any base, not just base 0, with almost identical proofs.] (5)* Suppose X, X 2,... are metric spaces, suppose d n is the metric on X n, and that d n (x, y) for all x, y X n. On the product n= X n define d({x n }, {y n }) = n= 2 n d n (x n, y n ), for {x n }, {y n } n= X n. (a) Prove that d is a metric on n= X n, (b) Prove that if each X n is complete, then so is n= X n (with the metric d. ) 3. Nonnegative Series, and tests for series convergence. A series k a k is called a nonnegative series if all the terms a k are 0.

9 INTRODUCTION TO REAL ANALYSIS II MATH 4332 BLECHER NOTES 9 For a nonnegative series, the sequence {s n } of the partial sums is a nondecreasing (or increasing) sequence. Indeed if s n = a 0 + a + + a n say, then s n+ = a 0 + a + + a n + a n, so that s n+ s n = a n 0. Therefore, by a fact we saw in 3333 for monotone sequences, the sum of the series equals the least upper bound of the sequence {s n } of partial sums. Thus the sum of the series always exists, but may be. More importantly, a nonnegative series converges if and only if the {s n } sequence is bounded above. The latter happens if and only if the sum of the series is finite. Thus to indicate that a nonnegative series converges we often simply write k a k <. Example: The HARMONIC SERIES is the important series = k= k. This is a nonnegative series, so to see if it converges we need only check if the sequence {s n } is bounded above, where s n = n. A trick to do this is to look at n+ x dx, interpreted as the shaded area in the graph below [Picture drawn in class]. This shaded area is less than the area of the n rectangles shown. Hence n So s n ln(n + ) ln() as n. n+ Thus the harmonic series diverges; it has sum +. x dx. The trick used in the previous example can be used in the same way to prove: The Integral Test: If f(x) is a continuous decreasing positive function defined on [, ) [Picture drawn in class], then k= f(k) converges if and only if f(x) dx converges (i.e. is finite). p-series. An almost identical argument shows that k= and only if p >. These are called p-series. Basic Comparison Test: Suppose that 0 a k b k for all k. ) If k b k converges, then k a k converges. 2) If k a k diverges, then k b k diverges. k p converges if [Proof: We have n k= a k n k= b k. So for ), if ( n k= b k) is bounded above then ( n k= a k) is bounded above. That is, by the third bullet in this section, if k b k converges, then k a k converges.

10 0 INTRODUCTION TO REAL ANALYSIS II MATH 4332 BLECHER NOTES Note that 2) is the contrapositive to ).] Limit Comparison Test: Suppose that k a k and k b k are nonnegative series. If lim sup If lim inf k a k b k k a k b k <, and if k b k converges then k a k converges. > 0 and k b k diverges then k a k diverges. [Proof: Suppose that s = lim sup k exists N with a k b k a k b k <. By (6) in Homework there < s + for k N. So a k < (s + )b k for k N. By the Basic Comparison Test, if k b k converges then k a k converges. If s > 0 then by the liminf variant of (6) in Homework there exists N with a k b k > s s 2 = s 2 for k N. So a k > s 2 b k for k N. By the Basic Comparison Test, if k b k diverges then k a k diverges.] Root Test: Suppose that k a k is a nonnegative series with lim sup (a k ) k = k r. If 0 r < then k a k converges. If < r then k a k diverges. [Proof: Suppose that r < c <. By (6) in Homework (with ɛ = c r) there exists N with a k k < c for k N. So a k < c k. Now k ck converges (geometric series), so by the Basic Comparison Test, k a k converges. a k k If < c < r then by (6) in Homework there are infinitely many k with Test.] > c, or equivalently, a k > c k >. So k a k diverges by the Divergence Ratio Test: Suppose that k a k is a nonnegative series with lim sup a k+ a k R and lim inf k then k a k diverges. k a k+ a k = = r. If 0 R < then k a k converges. If < r [Proof: Suppose that R < c <. By (6) in Homework (with ɛ = c r) there exists N with a k+ a k < c, for k N. So a N+ < ca N, a N+2 < ca N+ < c 2 a N, etc. Generally a N+k < c k a N. Now k ck a N converges (geometric series), so by the Basic Comparison Test, k a N+k converges. So k a k converges. A similar argument does the case r > c >. By the liminf variant of (6) in Homework there exists N with a k+ a k > c for k N. In this case a N+k > c k a N. Now c k a N as k, so by the Divergence test, k a N+k diverges. So k a k diverges.] From Homework 3 Question 3 (a) below it is easy to see that the root test is more powerful theoretically than the ratio test. That is if the ratio test works to prove convergence or divergence, then the root test would give the

11 INTRODUCTION TO REAL ANALYSIS II MATH 4332 BLECHER NOTES same conclusion. However the converse is not true (if the root test works, the ration test might not). See Homework 3 Question 2. Condensation test Suppose that a 0 a a 2 0, and that lim k a k = 0. Then k a k converges iff k 2k a 2 k converges. [Proof: Let s n = n k=0 a k and t n = n k=0 2k a 2 k. If n < 2 k then s n a + (a 2 + a 3 ) + (a a 7 ) + + (a 2 k + a 2k a 2 k+ ) a + 2a 2 + 4a k a 2 k = t k. Thus if (t k ) is bounded then (s n ) is bounded; and so by the fact in the paragraph before the harmonic series a few pages back, k a k converges if k 2k a 2 k converges. If n > 2 k then s n a + a 2 + (a 3 + a 4 ) + + (a 2 k a 2 k) 2 a + a 2 + 2a k a 2 k = 2 t k. So t k 2s n. Thus if (s n ) is bounded then (t k ) is bounded; and so by the fact in the paragraph before the harmonic series a few pages back, k 2k a 2 k converges if k a k converges. As an application of the condensation test note that the harmonic series n= n diverges because k 2k 2 k = k =. Homework 3 (due Thursday January 29). () Test for convergence (giving reasons): (a) (c) n= n n, (d) (f) n= +n 2 n n. n=2 (log n) log n, 2, (b) n n=2 (e) n= (n n ) n, n= n3 (log n) 2, (2) Suppose n= a n is the series a Find lim sup n+ a 3 n a n, lim inf n+ n a n and lim sup n a n. Can we conclude that n a n converges using the ratio test? Using the root test? (3) (a*) Show (or look up) that lim inf n s n+ /s n lim inf n s n /n lim sup n for any sequence {s n } in R \{0}. s n /n lim sup n s n+ /s n, (b) From (a) it is easy to see that the root test is more powerful theoretically than the ratio test. For example, deduce that if { s n+ /s n } converges then { s n /n } converges to the same limit. (c) Calculate lim n (n!) /n.

12 2 INTRODUCTION TO REAL ANALYSIS II MATH 4332 BLECHER NOTES 4. Absolute and conditional convergence A series k a k is called absolutely convergent if k a k converges. Recall that a k here could be a complex number (you can view complex numbers as elements of R 2 here). Any absolutely convergent series is convergent. [Proof: By the Cauchy test above, since k a k converges, given ɛ > 0 there exists an N 0 such that a n+ + a n a m a n+ + a n a m < ɛ, m > n N. By the Cauchy test again, k a k converges.] The converse is false, a series may be convergent, but not absolutely convergent. Such a series is called conditionally convergent. The Alternating Series Test (a.k.a. Leibniz Test)/Alternating Series approximation: Suppose that a 0 a a 2, and that lim k a k = 0. Then a 0 a + a 2 a 3 + (which in sigma notation is k=0 ( )k a k ) converges, and moreover s n k=0 ( )k a k a n for all n, where s n is the nth partial sum n k=0 ( )k a k. [Proof: We have noticed before that if m > n then s m s n = b n + b n+ + + b m, where b k is the kth term. Here b k = ( ) k a k, where a k is as above. Note that a n+k a n+k+ 0, so a n a n+ + a n+2 a n Hence s m s n = a n a n+ + a n+2 = a n (a n+ a n+2 ) (a n+3 a n+4 ) a n, since a n+k+ a n+k. It follows that (s n ) is Cauchy, so convergent. That is, k=0 ( )k a k converges. Letting m and using a fact about sequences from the prerequisite, we have k=0 ( )k a k s n a n.] Example. Approximate the sum of k= ( ) k+ k 4 with an error of less than Solution. The error in using s n to approximate the sum is < a n+ = (n+) 4 (note the starting index of this series is, not 0, which means we have to change the formula in the last result slightly). Now (n+) < if (n + ) 4 > 000. Choosing n = 5 will work. So an approximation with an error of less than 0.00 is s 5 = 5 k= ( ) k+ k 4 = (calculator). Exercise. Test for convergence/absolute or conditional convergence: ( ) n 2 n n= n!.

13 INTRODUCTION TO REAL ANALYSIS II MATH 4332 BLECHER NOTES 3 If k= a k converges absolutely then k= a k k= a k. [Proof. Take the limit as n in the inequality, nd using a fact about sequences from the prerequisite, we have n k= a k n k= a k.] If k= a k is a series, and f : N N is a bijection, then the series k= a f(k) is called a rearrangement of k= a k. It is not hard to see that a rearrangement of a convergent series need not converge. Theorem Any rearrangement of an absolutely convergent series is convergent and has the same sum. [Proof: Suppose that k= a k is absolutely convergent, and that ɛ > 0 is given. Then k= a k is convergent too, with sum s say, so s n s where s n = n k= a k. Choose N with k=n a k < ɛ 2 (see HW 2 Q ), and s n s < ɛ 2 for n N. Choose M N such that {, 2,, N} {f(), f(2),, f(m)} (this is possible since f is bijective why?). Let t n = n k= a f(k). If n > M we have that all the terms a k in s n = n k= a k with k N cancel with terms a f(j) in t n = n k= a f(k). So (and also using the triangle inequality), So s n t n a N+ + a N+2 + < ɛ 2. t n s t n s n + s n s < ɛ 2 + ɛ = ɛ, n M. 2 Hence t n s. So the rearrangement is convergent and has the same sum. ] Dirichlet test: Let k a k be a series whose partial sums form a bounded sequence. Suppose (b n ) is a decreasing sequence with limit 0. Then k a kb k converges. Abel s test: Suppose that k a k converges and b n is a monotonic convergent sequence. Then k a kb k converges. We will not prove the last two results (see e.g. Apostol s text for proofs, they are not hard). Homework 4 (due Tuesday February 3). () Let k= a k be a series, and suppose that n < n 2 < are integers. Let b = n k= a k, b 2 = n 2 k=n a + k, b 3 = n 3 k=n a 2+ k,. We call k= b k a series obtained from k= a k by adding parentheses. Prove that if k= a k converges then k= b k converges and has the same sum.

14 4 INTRODUCTION TO REAL ANALYSIS II MATH 4332 BLECHER NOTES Also: (a) If k= a k is a nonnegative series, prove that k= a k converges iff k= b k converges. (b)* If the sequence (n k+ n k ) is bounded and lim n a n = 0, show that k= a k converges iff k= b k converges. (2) Prove the fundamental fact about power series in the last bullet before Homework. (3) Suppose that z n = a n + ib n, where a n, b n R. Show that (a) convergent if and only if both a n and b n are convergent; (b) zn is zn is absolutely convergent if and only if both a n and b n are absolutely convergent; (c) ( ) ( ) n n= n + i n is convergent but not absolutely convergent. 2 (4) If n a n converges absolutely, show that n a2 n and a n n +a n converge absolutely (you may assume if you wish that that no a n = ). If n a n diverges show n n a n diverges. (5)* Prove or look up a proof of Riemann s result that any conditionally convergent series has the property that if a R is given, there is a rearrangement of the series which has sum a. Show also that there is a rearrangement of the series which diverges. 5. Double sums This section is really about interchanging double sums: that is when n= m= equals m= n=. If we have numbers a m,n 0 for all m, n N, define n,m= a m,n to be the supremum over all partial sums S N = N n,m= a m,n, for N N. Theorem 5.. If a m,n 0 for all m, n N, then the following sums are equal: ( ) ( ) a m,n = a m,n = a m,n. n= m= m= n= n,m= Proof. We leave this an exercise in sups of positive numbers, using the fact that the three double sums here are really just N M sup{ sup{ a m,n : M N} : N N}, n= m= M N sup{ sup{ a m,n : N N} : M N}, m= n= and sup{ N n= N m= a m,n : N N}. Next we allow the a m,n to be negative as well as positive; or even complex. Again we have partial sums S N = N n,m= a m,n, for N N. Now we say that

15 n,m INTRODUCTION TO REAL ANALYSIS II MATH 4332 BLECHER NOTES 5 a m,n converges if there is a number s such that for all ɛ > 0 there exists K such that s N M n= m= a m,n < ɛ whenever N K, and M K. We write this number s as n,m= sometimes simply as n,m a m,n. a m,n, or We remark that if a m,n 0 for all m, n N then saying that n,m a m,n converges is the same as saying that n,m= a m,n (in the sense defined before Theorem 5.) is finite. See Exercises below. Theorem 5.2. If a m,n are real (or complex) for all m, n N, and if n,m= a m,n <, then n,m a m,n converges, and all of the following sums are finite and we have: ( ) ( ) a m,n = a m,n = a m,n = lim S N. N n= m= m= n= n,m= Proof. We leave this as an exercise for the graduate students in the class (note that the last = is almost immediate by taking N = M in the definition of the sum n,m= a m,n). Double series obey the same rules as ordinary series. For example, a m,n + b m,n = (a m,n + b m,n ), ca m,n = c n,m= n,m= n,m= n,m= n,m= provided the first two double series converge, and c is a constant (scalar). a m,n, proofs are the same as before. We can write a convergent double series n,m= a m,n as an ordinary series. Indeed if g : N N N is a bijective function set b k = a g(k) and consider k b k. Theorem 5.3. If a m,n are real (or complex) for all m, n N, and if n,m= a m,n <, and b k is as defined above, then k b k converges (absolutely), and equals n,m= a m,n. The Proof. We can think of S N = N n,m= a m,n as the N-th partial sum of the series k c k where c = a,, and c 2 = a,2 + a 2,2 + a 2,, c 3 = a,3 + a 2,3 + a 3,3 + a 3, + a 3,2,. The associated series a, + a,2 + a 2,2 + a 2, + a,3 + a 2,3 + a 3,3 + a 3, + a 3,2 + a,4 + (note inserting parentheses in this series gives k c k) converges absolutely since a, + a,2 + a 2,2 + a 2, + a,3 +

16 6 INTRODUCTION TO REAL ANALYSIS II MATH 4332 BLECHER NOTES has partial sums dominated by n,m= a m,n <. So by Homework 4 Question, k c k converges. Moreover k b k is a rearrangement of the series in the second last centered formula. So by our earlier theorem on rearrangement of series, k b k converges (absolutely), and by Homework 4 Question its sum equals k c k = lim N S N = n,m= a m,n. We have also used here the fact about S N at the start of the proof. The last = uses a formula from the previous theorem. The Cauchy product of two series n=0 a n and n=0 b n is the series n=0 c n where n c n = a k b n k. k=0 A theorem of Mertens (see Apostol) says that if n=0 a n converges absolutely with sum s, and if n=0 b n converges with sum t, then the Cauchy product series converges and has sum st. A special case of this is found in Homework 5 Question 6 below. Homework 5 (due Thursday February 5). () Discuss the convergence of n,m nm n 2 +m 2 and n,m 2 (n2 +m 2). (2) Prove Theorem 5. in detail. (3) If a m,n 0 for all m, n N show that n,m a m,n converges iff n,m= a m,n (in the sense defined before Theorem 5.) is finite. (4)* Prove Theorem 5.2. (5) If n a n and n b n are absolutely convergent series with sums s and t respectively, show that n,m a n b m is an absolutely convergent double series whose sum is st. (6) If n=0 a n and n=0 b n both converge absolutely, show that the Cauchy product series converges (absolutely) and has sum st. [Hint: This follows from Homework 5 Question 5 and Theorem 5.3, and Homework 4 Question, because the Cauchy product series is one of the kind considered in Theorem 5.3, but then with parentheses added as in Homework 4 Question.] (7) If s n is the nth partial sum of a series k a k, define the Cesáro means to be the sequence (σ n ) defined by σ n = n (s + s s n ). We say that k a k is Cesáro summable if the sequence (σ n ) converges, and in this case lim n σ n is called the Cesáro sum of the series k a k. Prove that if k a k converges with sum s then it is Cesáro summable and its Cesáro sum is s. END OF CHAPTER 0