Math 316, Intro to Analysis Absolute convergence, conditional convergence, reordering series. Recall that the comparison tests work only for series of positive terms. Today we deal (very slightly) with series which have some negative terms. An interesting thing can happen: A series might converge if you allow for negative terms to cancel with positive ones, but when you make all of the terms positive, the same series may diverge!
∞ X Definition 1. A series ak is called absolutely convergent if . k=1 ∞ X The series is called conditionally convergent if ak converges but . k=1 Example 2. At the end of these notes (which we may not make it to) we will see that for 1 if n = 2k − 1 is odd 1 −1 1 −1 k an = 1, −1, , , , ,... = −1 2 2 3 3 if n = 2k is even k ∞ ∞ ∞ X X X while ak converges, |ak| diverges. Thus, ak is conditionally convergent. k=1 k=1 k=1 The following proposition reveals that absolute convergence is a stronger condition than is conditional convergence. ∞ X Proposition 3 (Absolute convergence implies convergence). If |ak| converges then so k=1 ∞ X does ak. k=1 Proof. The proof will pass through the Cauchy criterion. ∞ X Let (Sn) be the sequence of partial sums for ak and (Tn) be the sequence of partial k=1 ∞ ∞ ∞ X X X sums for |ak|. Suppose that |ak| converges. To prove that ak converges, we must k=1 k=1 k=1 prove that converges or equivalently that it is Cauchy. Consider any > 0. We must show:
Goal:
∞ X Since |ak| converges by assumption, it must be that the sequence ( ) converges k=1 and so is Cauchy. Since we have already assumed > 0, we see that there exists an N such that
Now consider the N above and consider any m > n > N. Using the triangle inequality, 1 2
|Sm − Sn| =
To summarize, for any > 0 we have found an N such that if m, n > N then |Sm −Sn| < . ∞ X (Sn) is Cauchy and so converges. Thus, an converges. k=1 We begin by revealing how to detect conditionally convergent series: ∞ X Theorem 4. Suppose ak is a conditionally convergent sequence. Define two new se- k=1 quences bn = max(an, 0) (The positive terms) and cn = min(an, 0) (the negative terms). ∞ ∞ X X Then bk and ck each diverge. k=1 k=1 Example 5. Consider the (allegedly) conditionally convergent sequence 1 if n = 2k − 1 is odd k an = −1 if n = 2k is even k ∞ X of Example 2. Since ak is conditionally convergent it must be that for k=1
bn = max(an, 0) =
∞ ∞ X X 1 the series b is . This can be used to show that is . k k k=1 k=1 ∞ X Proof of Theorem 4. Let ak be a conditionally convergent series. For all k, let bk = k=1 max(ak, 0) and ck = min(ak, 0). We begin by noticing that for all k, ak = bk + ck. (If you want to prove this, make a case by case proof depending on the sign of ak.) ∞ ∞ X X Suppose for the sake of contradiction that bk converges. By assumption ak con- k=1 k=1 ∞ ∞ ∞ ∞ X X X X verges. Then ck = = − is a series by k=1 k=1 k=1 k=1 limit laws. (Theorem 2 from Monday.) ∞ ∞ X X By design, bk ≥ 0 for all k. Thus |bk| = bk is , k=1 k=1 3
∞ ∞ X X Similarly, ck ≤ 0 for all k. Thus, |ck| = −ck is . k=1 k=1 ∞ ∞ ∞ X X X But then |ak| = |bk + ck| (|bk| + |ck|) = . By the com- k=1 k=1 k=1 ∞ X parison test we see that |ak| is . This contradicts the assumption of k=1 conditional convergence.
Theorem 6 (The alternating series test). If bn ≥ 0 is any non-negative decreasing sequence ∞ X k+1 with bn → 0 then (−1) bk converges. k=1 Example 7. Consider the (allegedly) conditionally convergent sequence 1 if n = 2k − 1 is odd k an = −1 if n = 2k is even k ∞ X can you use this theorem to prove that ak converges? k=1
Proof of Theorem 6. Assume that bn is positive, decreasing and converges to zero. Let Tn = n X k+1 (−1) · bk be the corresponding sequence of partial sums. Then: k=1 n X T2n = b2k−1 − b2k k=1
Since (bn) is each b2k−1 − b2k 0, so that the sequence T2n is . If we can show then we will be able to use monotone convergence to get T2n to converge. Writing T2n slightly differently:
n ! X T2n = b1 − b2k − b2k+1 − b2n k=1 Since b2k − b2k+1 0 and b2n 0 we see that T2n ≤ b1. T2n is and and so converges. Let L be the limit of T2n. We claim that L is the limit of Tn. Consider any > 0 then there is an M such that for all m > M |T2m−L| < and |bm| < . Let N = 2M + 1 and consider any n > N. If n = 2m is even then since m > M we have that |T2m − L| < . If n = 2m + 1 then by the triangle inequality and taking advantage of the fact that m > M,
|T2m+1 − L| = |T2m − L + b2m+1| ≤
This completes the proof. 4
∞ X Use the alternating series test to show that (−1)k+1/k is conditionally convergent. (You k=1 ∞ X may assume that 1/k is divergent.) k=1
Absolutely convergent sequences can be reordered.
In the case of finite sums it does not matter what the order of summation is. 2n n n X X X bk is the same as b2k + b2k−1. k=1 k=1 k=1 Is the same true for infinite sums? Can we re-arrange terms of an infinite sum willy nilly? First let’s figure out what it means to re-order the terms of a sequence: Let an be a sequence (recall that this is just a code for a function from to .)
Let σ : N → N be a bijection. Then one can define a new sequence by composition: bn = aσ(n). 1 if n = 2k − 1 is odd k Think about the sequence from example 2. Let an = −1 . if n = 2k is even k Take the bijection which re-orders by listing two evens for every odd: N 2(2k − 1) if n = 3k − 2 σ(n) = (2, 4, 1, 6, 8, 3, 10, 12, 5) = 4k if n = 3k − 1 . Let bn = aσ(n). 2k + 1 if n = 3k ∞ ∞ X X We will compute an and bn. Remarkably, they will be different! k=1 k=1 Before re-ordering: Compute S2n by reordering the finite sum: 2n n n n n X X X X X S2n = ak = a2k + a2k−1 = + = k=1 k=1 k=1 k=1 k=1
On the other hand S2n−1 = S2n − a2n =
∞ X So that ak = lim Sn = k=1 5
after re-ordering. Get an explicit formula for bn = aσ(n):
b3k−2 = a2(2k−1) =
b3k−1 = a4k =
b3k = a2k−1 = Making these substitutions and reordering a finite sum, 3n X S3n = bk k=1
Notice that every term of this final sum is negative: The sub-sequence S3n is . −1 In particular: S3n S3 = 2 . ∞ X Thus, if bk even converges, it must be to a negative number! k=1 ∞ X A tangential question: Does −bk even converge? k=1 So reordering infinite sums so that negative terms start appearing more (or less) frequently ∞ ∞ X X can change the value of the sum! It is worth noting that while ak = 0 converges |ak| k=1 k=1 diverges.
∞ X Theorem 8. If ak is absolutely convergent and converges to L, then for any reordering k=1 ∞ X bn = aσ(n), bk = L. k=1
Notice that as a consequence of this theorem and the starting example we see that our initial series is conditionally convergent.
1 if n = 2k − 1 is odd ∞ k X Claim 9. For an = −1 the series ak is conditionally conver- if n = 2k is even k k=1 gent. 6
1 if n = 2k − 1 is odd k ∞ Proof. Let an = −1 . We have already proven that Sumk=1ak = if n = 2k is even k . In particular the series converges. ∞ X Now suppose for the sake of contradiction that ak converged absolutely, then by The- k=1 ∞ X orem 8 if (bn) were any reordering of (an) it would follow that bk = . But we k=1 ∞ X produced a reordering for which bk , it it converges at all. k=1
It is remarkable that conditionally convergent sequences fail to have this property very badly. One can reorder conditionally convergent sequences to get ANYTHING. The proof of the following theorem will be an extra credit exercise.
∞ X Theorem 10. If ak is conditionally convergent then for any number L there exists a k=1 ∞ X reordering bn = aσ(n) for which ak = L. k=1
Remark 11. The reordered series converging to L could have been replaced by divergence or divergence to ±∞ and the theorem would still hold. Reordering conditionally convergent series is ridiculously violent.
The remainder of today’s lecture amount to proving Theorem 8.
X Proof of Theorem 8. Let ak be absolutely convergent. Let bn = aσ(n) be any reordering. X Since ak is absolutely convergent, there is an N1 such that for all m > n > N1, n X |ak| < . k=m
∞ n X X On the other hand, since ak = L there is an N2 such that for all n > N2, ak − L < k=1 k=1 . Let M = max(N1,N2) Since σ : N → N is onto there is an N such that σ[{1, 2,...,N}] ⊇ {1, 2, 3,...,M}. Let n > N and X = σ[{1, 2, . . . , n}] − {1, 2, 3,...,M}. Then by reordering the finite sum and using the triangle inequality
! n n M X X X X bn − L = aσ(n) − L = ak − L + ak ≤ k=1 k=1 k=1 k∈X 7
M X Notice that since M > N2, ak − L < . Now we deal with the sum k=1 involving X. Using the triangle inequality and setting ourselves up to take advantage of absolute convergence.
X ak ≤ k∈X
On the other hand, since we removed {1, 2,...,M} from {σ(1), . . . , σ(n)} to get X we have that X ⊆ {M + 1,M + 2,... max(X)}. Removing some terms from a sum of non-negative terms reduces the value: X |ak| ≤ k∈X
Now since N1 < M + 1 < max(X), we see that
X ak ≤ k∈X
n X and so bn − L ≤ k=1
Which completes the proof.