Proving a Sequence Converges - Cauchy?

Question

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.

Answer 1

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 $(1).$

Answer 2

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 $\{a_n\}$.

Answer 3

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

Answer 4

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.