# Proving a Sequence Converges - Cauchy?

Let $a_n$ be q bounded sequence such that $$a_n-a_{n+1}\leq \frac{1}{2^n}$$ Let b_n be a sequence such that $$b_n=a_n-\frac{1}{2^{n-1}}$$

1. Prove that $a_n$ converges.
2. Prove that $b_n$ converges.

I think it's quite clear that if I prove (1), (2) will be immediate, from limit arithmetic.

I think I should solve this with Cauchy Sequences, I'm just not sure how.

• Actually do you have $a_n - a_{n+1} \le 2^{-n}$ or $|a_n - a_{n+1}|\le 2^{-n}$ or $0 \le a_n - a_{n+1} \le 2^{-n}$? I don't think this works as giving. I'magine $a_n-a_{n+1}$ being "large" negative values sometimes. Say for $n$ odd we have $a_n = n$ for $a_{n+1} = n - 2^{-n}$. That doesn't converge but if $n is odd$a_n - a_{n+1} = 2^{-n}$and if$n$is even$a_n - a_{n+1} = 2^{-{n-1}} - 1 < 0 < 2^{-n}$. – fleablood Apr 12 '17 at 19:39 • @fleablood the question I have doesn't have absolute values, I double checked... – Alan Apr 12 '17 at 19:41 • I was also curious about this issue - what if some$n$'s are negative, and "jump" between values? – Alan Apr 12 '17 at 19:41 • Counter example then$a_{2n - 1} = 2n - 1$and$a_{2n} = 2n-1 - 2^{-n}$. Clearly doesn't converge, and if I did my math right$a_n - a_{n+1}$is either negative ($2^{-k} - 1$) or a very small positie$2^{-k}$. I maybe didn't do my math right but it can be fixed if I didn't. – fleablood Apr 12 '17 at 19:46 ## 4 Answers The conclusion in Problem 1 is correct as stated; proof below. (I can't understand why it is coupled with problem 2, which is pretty trivial.) Since$a_n$is bounded,$\limsup a_n$is a finite number$L.$Claim:$\lim a_n = L.$Proof: Let$\epsilon>0.$Properties of the$\limsup$imply that there exist infinitely many$N$such that $$\tag 1|a_N-L|<\epsilon/2\,\, \text { and } a_n < L+\epsilon,n\ge N.$$ By taking one of these$N$large enough we will also have$\sum_{N}^{\infty} \frac{1}{2^n} < \epsilon/2.$So now fix such an$N.$Then we have $$a_{N+1}\ge a_N - 1/2^N,$$ and $$a_{N+2}\ge a_{N+1} - 1/2^{N+1} \ge a_N - (1/2^N+1/2^{N+1}),$$ and so on. The general situation is then $$a_{N+k} \ge a_N - \sum_{n=N}^{N+k-1}\frac{1}{2^n} > a_N - \epsilon/2> L-\epsilon.$$ Thus$a_n > L-\epsilon$for$n\ge N.$The claim follows from this and the inequality on the right of$(1).$• Note that the$1/2^n$is completely arbitrary. All that is required is that the$a_n$be bounded above and$a_n-a_{n+1}\le b_n$where$\sum b_n$converges. I gave you an homage in this answer, which is a generalization of the solution you provided herein. -Mark – Mark Viola Apr 14 '17 at 4:56 • You're welcome. And (+1) for this nice development. I owe you one - or more. – Mark Viola Apr 14 '17 at 17:08 Hint. By telescoping, one has $$a_N=(a_N-a_{N-1})+(a_{N-1}-a_{N-2})+\cdots+(a_1-a_0)+a_0$$ Can you use it? Then the convergence of$\{b_n\}$follows from the convergence of$\{a_n\}$. • Thanks. Does it involve using$\Sigma$? – Alan Apr 12 '17 at 19:25 • You are welcome. You can use either$\sum$either the 'extensive' sum as in my hint above. – Olivier Oloa Apr 12 '17 at 19:27 • It does but$\sum\limits_{n=1}^k \frac 1{2^n} = 1 - \frac 1{2^{k}}$and$\sum\limits_{n=1}^{\infty} \frac 1{2^n} = 1$so$\sum\limits_{n=k+1}^{m} \frac 1{2^n} < \sum\limits_{n=k+1}^{\infty} \frac 1{2^n} = 1 - (1-\frac 1{2^k}) = \frac 1{2^k}$. – fleablood Apr 12 '17 at 20:05 Use the fact that if$m > n$then $$|a_m - a_n| \le \sum_{k=n}^{m-1} |a_{k+1} - a_k| \le \sum_{k=n}^{m-1} \frac 1{2^k} < \frac{1}{2^{n-1}}.$$ • Where did this come from, there are no absolute values mentioned. – zhw. Apr 12 '17 at 21:27 This turned out to be harder than I thought it would. Let$\epsilon > \frac 1{2^n} > 0$and let$b = \sup\{a_i; i \ge n+2\} \le \sup \{a_i\}$which exists as$\{a_i\}$is bounded. There exists and$a_K; K \ge n+2$so that$b-a_K < \epsilon/4$. [Note:$\frac 1{2^K} \le \frac 1{2^{n+2}} = \frac 14*\frac 1{2^n} < \epsilon/4$.] Now let's let$j > K$then if$a_j \ge a_K$then$a_K \le a_j \le b$and$|b - a_j|\le |b-a_K| < \epsilon/4 < \epsilon/2$. If$a_j < a_K$then $$0 < a_K - a_j = (a_K - a_{K+1}) + (A_{K+1} - A_{K+2}) + .... + (A_{j-1} - A_j) \le \sum_{n=K}^{j-1} \frac 1{2^n} = \frac 1{2^K}\sum_{n=0}^{j-K -1}\frac 1{2^n} \le \frac 1{2^K} < \epsilon/4$$. So$|b- a_j| = (b-a_K) - (a_K - a_j) < \epsilon/4 + \epsilon/4 = \epsilon/2$. So if$j,k > K$than$|a_j-a_k|\le |b-a_j| + |b-a_k| < \epsilon/2 + \epsilon/2 < \epsilon$So$\{a_i\}$is Cauchy. • In the OP it is assumed that$\{a_n\}\$ is bounded. – Olivier Oloa Apr 12 '17 at 20:16
• Yeah,... I just realized that. Hmm... – fleablood Apr 12 '17 at 20:20