Proving of an inequality of a sequence The question I am stuck on is that there is a sequence $(x_n)$ with
$$\forall n \in \mathbb{N}: |x_{n+1} - x_n| \leq \frac{1}{n(n+1)} $$
It asks to prove that 
$$\forall m \in \mathbb{N}:\forall n \in \mathbb{N}: |x_{m} - x_n| \leq |\frac{1}{m} - \frac{1}{n}| $$
I thought of using the triangle inequality and induction, but I'm not sure how to do it. 
Part 2 asks to prove that the sequence $(x_n)$ is convergent.
The question was from a past exam paper that I am using to study for my upcoming exam. Thanks in advance for any help.
 A: Triangle inequality and induction is definitely one way to go. Fix $n\in \mathbb{N}$, Then 
$$
\vert x_{n+1}-x_n\vert\leq \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\;, 
$$
so the claim is true for $n+1$. This is the base case of the induction for fixed $n$. Now assume the claim is true for $m>n$, we have
\begin{eqnarray}
\vert x_{m+1}-x_n \vert=\vert x_{m+1}-x_m+x_m-x_n \vert &\leq& \vert x_{m+1}-x_m\vert+ \vert x_m-x_n \vert
\\
&\leq& \frac{1}{m}-\frac{1}{m+1}+\frac{1}{n}-\frac{1}{m}
\\
&=&\frac{1}{n}-\frac{1}{m+1}
\end{eqnarray}
So by induction the claim is true for all $m>n$. Since $n$ was arbitrary, the claim is true for all $m$ and $n$. 
For part 2, note that the sequence is Cauchy. Indeed, let $\epsilon>0$ and fix $N$ such that $\frac{2}{N}<\epsilon$. Then for all $m,n>N$
$$
\vert x_m-x_n\vert\leq \vert \frac{1}{n}-\frac{1}{m}\vert\leq \frac{2}{N}<\epsilon
$$
Assuming your in a complete metric space (like $\mathbb{R}$) the sequence must be convergent.
A: Just a hint: Assuming $m>n$ use a telescopic sum to write 
$$|x_m-x_n| = | \sum_{i=n}^{m-1} (x_{i+1}-x_i) |$$
Tell me if you need more help ;)
