Proving that a sequence converges if $|a_n - a_{n+1}| < Mr^n$ for some $M > 0$ and $r \in (0,1).$

Question

Let $\{a_n\}_n$ be a sequence. Suppose that there exist $M \gt 0$ and $r \in (0, 1)$ such that $|a_n - a_{n+1}| \lt Mr^n$ for all $ n \in \Bbb N.$ Prove that $\{a_n\}_n$ converges. I'm not really sure where to go about with this problem. I feel like the easiest way would be to prove that the sequence is Cauchy, but I'm not entirely sure. Any help would be appreciated, thanks!

Answer 1

You may start by considering the following set of inequalities: $$ \begin{align} |x_{n+1} - x_n| &< Mr^n\\ |x_{n+2} - x_{n+1}| &< Mr^{n+1}\\ |x_{n+3} - x_{n+2}| &< Mr^{n+2}\\ &\cdots \\ |x_{n+p+1} - x_{n+p}| &< Mr^{n+p} \end{align} $$

Sum up the inequalities to obtain: $$ |x_{n+1} - x_n| + |x_{n+2} - x_{n+1}| + \cdots + |x_{n+p+1} - x_{n+p}| < M\sum_{k=0}^pr^{n+k} $$

By geometric sum: $$ M\sum_{k=0}^pr^{n+k} = M\frac{r^n(1-r^{p+1})}{1-r} \le M\frac{r^n}{1-r} $$

You can now choose some $\epsilon > 0$ and $N \in \Bbb N$ such that: $$ M\frac{r^N}{1-r} < \epsilon \tag1 $$

Define: $$ r = \frac{1}{1+q}, q\in\Bbb R_{>0} $$

Thus: $$ M\frac{r^N}{1-r} = M\frac{1}{(1-r)(1+q)^N} < \epsilon $$

Or: $$ N > \log_{1+q}\frac{M}{(1-r)\epsilon} \implies M\frac{r^N}{1-r} < \epsilon $$

Now going back to the sum of absolute values and using triangular inequality one may obtain: $$ |x_{n+1} - x_n + x_{n+2} - x_{n+1} + \cdots + x_{n+p+1} - x_{n+p}| \le |x_{n+1} - x_n| + |x_{n+2} - x_{n+1}| + \cdots + |x_{n+p+1} - x_{n+p}| $$

Note the telescoping nature and you get: $$ |x_{n+1} - x_{n+p+1}| < M\frac{r^n}{1-r} < \epsilon $$

Now for all $n$ such that: $$ n > N > \log_{1+q}\frac{M}{(1-r)\epsilon} $$

the inequality $(1)$ holds, which shows that the sequence satisfies Cauchy's criteria hence convergent.

Answer 2

Hint: for $m > n$, $$|a_n - a_m|\le M(r^n + r^{n+1} +\cdots + r^{m-1})$$ (sum of a geometric progression $=\cdots$ ?)

Answer 3

We have $$\sum_{n=1}^\infty |a_{n+1}-a_n| \le \sum_{n=1}^\infty Mr^n = \frac{Mr}{1-r} < +\infty$$ Absolute convergence of a series implies convergence so $$a_1 + \sum_{n=1}^\infty(a_{n+1} - a_n) = a_1 + \lim_{N\to\infty} \sum_{n=1}^{N-1}(a_{n+1} - a_n) = a_1 + \lim_{N\to\infty} (a_N - a_1) = \lim_{N\to\infty} a_N$$ also exists.