Exercise 3.15

Exercise 15: Definition 3.21 can be extended to the case in which the a n lie in some fixed R k . Absolute convergence is defined as convergence of | a n | . Show that Theorems 3.22, 3.23, 3.25(a), 3.33, 3.34, 3.42, 3.45, 3.47, and 3.55 are true in this more general setting.

Answers

Theorem 3.22

For a n k , the series a n converges if and only if for every 𝜖 > 0 there is an integer N such that

| k = n m a k | 𝜖

for all m n N .

Proof: (→) Suppose that a n converges. Then by definition the sequence { s n } of partial sums converges where s n = k = 1 n a k . It follows from Theorem 3.11a that { s n } is a Cauchy sequence, noting that this is true in any metric space. So consider any 𝜖 > 0 . Then, since { s n } is Cauchy, there is an integer N such that for every m N and n N , | s m s n | < 𝜖 .

So let M = N + 1 and consider any m n M . Also let l = n 1 , from which it follows immdediately that n = l + 1 . Clearly then m l N so that | s m s l | < 𝜖 . We then have

| k = n m a k | = | k = l + 1 m a k | = | k = 1 m a k k = 1 l a k | = | s m s l | 𝜖

as required.

(←) Suppose that for every 𝜖 > 0 there is an integer N such that

| k = n m a k | 𝜖

for all m n N and consider any 𝜖 > 0 . Then obviously there is an integer N for which

| k = n m a k | 𝜖 2

for every m n N . So consider any m N and n N . We can assume that m n without loss of generality. If m = n then we have simply | s m s n | = | s m s m | = | 0 | = 0 < 𝜖 . If m > n let l = n + 1 . It then follows that m l N so that

| k = l m a k | 𝜖 2 .

We then have

| s m s n | = | k = 1 m a k k = 1 n a k | = | k = n + 1 m a k | = | k = l m a k | 𝜖 2 < 𝜖 .

Thus we have shown that { s n } is a Cauchy sequence. Then by Theorem 3.11c, { s n } converges since we are in k . It follows then by definition that a n converges.

Theorem 3.23

For a n k , if a n converges then lim n a n = 0 .

Proof: Suppose that a n converges and consider any 𝜖 > 0 . Then by Theorem 3.22 there is an integer N such that

| k = n m a k | 𝜖 2

for all m n N . So consider any n N and let m = n . Then clearly m n N is true so that

| a n 0 | = | a n | = | k = n n a k | = | k = n m a k | 𝜖 2 < 𝜖 ,

thereby showing by definition that lim n a n = 0 .

Theorem 3.25a

For a n k and c n , if | a n | c n for all n N 0 where N 0 is a fixed integer, and if c n converges, then a n converges.

Proof: The proof is the same as that in the text since Theorem 3.22 applies to sums in k and the triangle inequality is also true there.

Theorem 3.33

For a n k , given a n , put α = limsup n | a n | n . Then

(a) if α < 1 then a n converges
(b) if α > 1 then a n diverges
(c) if α = 1 , then the test gives no information.

Proof: The proof is again the same as that given in the text since for each part:

(a) The comparison test (Theorem 3.25a) is valid in k .
(b) Theorem 3.23 is valid in k .
(c) The series given still provide a counter example since they are in 1

Theorem 3.34

For a n k , the series a n

(a) converges if limsup n | a n + 1 a n | < 1
(b) diverges if | a n + 1 a n | 1 for all n N 0 , where N 0 is some fixed integer.

Proof: Similar to the root test, the proof is the same as that given in the text since for each part:

(a) The comparison test (Theorem 3.25a) is valid in k .
(b) Theorem 3.23 is valid in k .

Theorem 3.42

For a n k and b n suppose that

(a) the partial sums A n of a n form a bounded sequence;
(b) b 0 b 1 b 2 ;
(c) lim n b n = 0 .

Then a n b n converges.

Proof: The proof given in the text still holds since Theorem 3.41 clearly holds for a n k as does the Cauchy criterion (Theorem 3.22).

Theorem 3.45

For a n k , if a n converges absolutely, then a n converges.

Proof: The proof is the same since the Cauchy criterion (Theorem 3.22) and the triangle inequality both hold in k .

Theorem 3.47

For a n k and b n k , if a n = A and b n = B , then ( a n + b n ) = A + B , and c a n = cA for any fixed c .

Proof: The proofs are the same as those given in the text since the limit rules used (Theorem 3.3 parts a and b) are trivially shown to hold in k .

Theorem 3.55

If a n is a series of elements in k that converges absolutely, then every rearrangement of a n converges, and they all converge to the same sum.

Proof: The proof is the same since the Cauchy criterion (Theorem 3.22) holds in k .

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2023-08-07 00:00
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