TITLE: how to show $f(1/n)$ is convergent?
QUESTION [3 upvotes]: Let $f:(0,\infty)\rightarrow \mathbb{R}$ be differentiable, $\lvert f'(x)\rvert<1 \forall x$. We need to show that
$a_n=f(1/n)$ is convergent. Well, it just converges to $f(0)$ as $\lim_{n\rightarrow \infty}f(1/n)=f(0)$ am I right? But $f$ is not defined at $0$ and I am not able to apply the fact $\lvert f'\rvert < 1$. Please give me hints.

REPLY [3 votes]: The condition $|f'(x)|<1$ implies that f is lipschitz. 
$$|f(x)-f(y)|\le |x-y|                                      $$
Then $$|f(\frac{1}{n})-f(\frac{1}{m})|\le|\frac{1}{n}-\frac{1}{m}|$$
Since $x_n=\frac{1}{n}$ is Cauchy, $f(\frac{1}{n})$ also is Cauchy
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