how to show $f(1/n)$ is convergent? 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.

By the mean value theorem, for every distinct $x>0$ and $y>0$, there exists $z>0$ between $x$ and $y$ such that
$$
f(x)-f(y)=f'(z)(x-y).
$$
It follows that $|f(x)-f(y)|\leq |x-y|$ for all $x>0$ and $y>0$.
Now recall that in $\mathbb{R}$ (which is complete), a sequence converges if and only if it is a Cauchy sequence.
The sequence $(1/n)$ converges to $0$, so it is a Cauchy sequence.
By the inequality above, it follows that $(f(1/n))$ is a Cauchy sequence, so it converges in $\mathbb{R}$.
