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.
 A: The function $f$ is Lipschitz continuous (since $f'(x)<1$) thus uniformly continuous. Therefore it sends Cauchy sequences to Cauchy sequences.
A: $f(0)$ is not yet well-defined, we have to be sure it exists. As the derivative is locally integrable, we can write 
$$|a_{m+n}-a_n|\leqslant\int_{\frac 1{m+n}}^{\frac 1n}|f'(t)|dt\leqslant \frac 1n-\frac 1{m+n}\leqslant \frac 1n,$$
so the sequence $\{a_n\}$ is Cauchy. 
A: 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
A: You are taking the limit as $x \rightarrow 0$, which means that you are not evaluating $f$ at 0, but in a neighborhood to the right of $x=0$. Thus the inequality applies.
A: 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 $\mathbb{R}$.
