Proving of an inequality of a sequence

Question

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.

Answer 1

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 ;)

Answer 2

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.