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.

  • $\begingroup$ 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 $a_n - a_{n+1} = 2^{-{n-1}} - 1 < 0 < 2^{-n}$. $\endgroup$ – fleablood Apr 12 '17 at 19:39
  • $\begingroup$ @fleablood the question I have doesn't have absolute values, I double checked... $\endgroup$ – Alan Apr 12 '17 at 19:41
  • $\begingroup$ I was also curious about this issue - what if some $n$'s are negative, and "jump" between values? $\endgroup$ – Alan Apr 12 '17 at 19:41
  • $\begingroup$ 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. $\endgroup$ – fleablood Apr 12 '17 at 19:46

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 $(1).$

  • $\begingroup$ 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 $\endgroup$ – Mark Viola Apr 14 '17 at 4:56
  • $\begingroup$ You're welcome. And (+1) for this nice development. I owe you one - or more. $\endgroup$ – 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 $\{a_n\}$.

  • $\begingroup$ Thanks. Does it involve using $\Sigma$? $\endgroup$ – Alan Apr 12 '17 at 19:25
  • $\begingroup$ You are welcome. You can use either $\sum$ either the 'extensive' sum as in my hint above. $\endgroup$ – Olivier Oloa Apr 12 '17 at 19:27
  • $\begingroup$ 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 $\sum\limits_{n=k+1}^{m} \frac 1{2^n} < \sum\limits_{n=k+1}^{\infty} \frac 1{2^n} = 1 - (1-\frac 1{2^k}) = \frac 1{2^k}$. $\endgroup$ – fleablood Apr 12 '17 at 20:05

Use the fact that if $m > n$ then $$|a_m - a_n| \le \sum_{k=n}^{m-1} |a_{k+1} - a_k| \le \sum_{k=n}^{m-1} \frac 1{2^k} < \frac{1}{2^{n-1}}.$$

  • $\begingroup$ Where did this come from, there are no absolute values mentioned. $\endgroup$ – 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.

  • $\begingroup$ In the OP it is assumed that $\{a_n\}$ is bounded. $\endgroup$ – Olivier Oloa Apr 12 '17 at 20:16
  • $\begingroup$ Yeah,... I just realized that. Hmm... $\endgroup$ – fleablood Apr 12 '17 at 20:20

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