If, $\forall n\in\mathbb{N}$, $\lvert x_{n+1} - x_n \lvert\le a^n$ (for $a\in (0,1)$), then $x_n\to x$

Question

Consider $\{x_n\}$, a sequence in $\mathbb{R}$, and suppose that $\exists a \in (0,1)$ such that $\lvert x_{n+1} -x_n \lvert\le a^n$ for all $n\in \mathbb{N}$. Then $x_n \to x$ for some $x\in \mathbb{R}$. $\leftarrow$ This is what needs to be proved.

I'm feeling somewhat puzzled because there's a counterexample, where $x_n = \log n$. So, we know that $0 <a^n < 1 $ for all $n\in\mathbb{N}$, but isn't $\lvert\log(n+1)-\log(n)\lvert \le a^n$ for all $n\in\mathbb{N}$, while $\{\log(n)\}$ diverges to $+\infty$?

Answer 1

$$|x_n-x_{n+m}|\leq |x_n-x_{n+1}|+|x_{n+1}-x_{n+2}|+...+|x_{n+m-1}-x_{n+m}|$$ $$\leq a^n+a^{n+1}+...+a^{n+m-1}=a^n(1+a+..+a^{m-1})\leq a^n(1+a+a^2+...)=a^n\frac{1}{1-a}\to 0,\,\forall m\ge 0$$ Therefore $\{x_n\}$ is Cauchy and thus converges.

Answer 2

Assume that $n<m$. Then by triangle inequality and by your condition \begin{align*} |x_m - x_n|&\leq \sum_{k=n}^{m-1}|x_{k+1}-x_k|\leq \sum_{k=n}^{m-1}a^{k}= \sum_{k=0}^{m-1}a^{k}-\sum_{k=0}^{n-1}a^{k}=\frac{1-a^{m}}{1-a}-\frac{1-a^{n}}{1-a}=\frac{a^{n}-a^{m}}{1-a}. \end{align*} Can you show that this implies that $(x_n)$ is a Cauchy sequence in $\mathbb{R}$? Thus has a limit?

Also, regarding your counter-example, note that \begin{align*} |\log(n+1)-\log(n)|&=|\log(1+\frac{1}{n})|\to 0 \end{align*} as $n\to \infty$, but the decay is not exponential. The series expansion of $\ln$ for $|x|\leq 1$ is \begin{align*} \ln(1+x)&=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}+O(x^4). \end{align*}