Prove that if $|x_{n+1} - x_n| \leq \frac{1}{n^2}$ for all terms of a sequence then the sequence is Cauchy. Prove that if $|x_{n+1} - x_n| \leq \frac{1}{n^2}$ for all terms of a sequence, $n \in \mathbb{N}$ then the sequence is Cauchy.
I have a proof but I'm not sure that it is correct, I would also like to know if there is a more obvious way to prove it.
I have $|x_{n+1} - x_n| \leq \frac{1}{n^2}$, so $|x_{n+2} - x_{n+1}| \leq \frac{1}{(n+1)^2}$. This follows that $|x_{n+2} - x_n| \leq \frac{1}{n^2} + \frac{1}{(n+1)^2}$. This can be generalised for any $i \in \mathbb{N}$, where
$|x_{n+i} - x_n| \leq \sum\limits_{n=1}^i\frac{1}{n^2}$
Now $|x_{n+2} - x_{n+1}| \leq \frac{1}{(n+1)^2}$, so $|x_{n+3} - x_{n+1}| \leq \frac{1}{(n+1)^2} + \frac{1}{(n+2)^2}$ through the same reasoning. Following this, we can generalise the inequality further for any $i,j \in \mathbb{N}$ to
$|x_{n+i} - x_{n+j}| \leq \sum\limits_{n=i}^j\frac{1}{n^2}$. The series $\frac{1}{n^2}$ converges to $\frac{\pi^2}{6}$, so for all $i,j \in \mathbb{N}$ I have that $|x_{n+i} - x_{n+j}| \leq \frac{\pi^2}{6}$. Thus the sequence is Cauchy.
 A: In order for the sequence to be Cauchy you need $|x_{n}-x_m|\to 0$ as $n,m\to \infty$, what you obtained is that this difference is bounded. However, let us call
$$
S_n:=\sum_{k=1}^{n} \frac{1}{k^2}.
$$
We have that $S_n$ is a sequence of real numbers that converges (to $\pi^2/6$, as you said), and therefore it is Cauchy, since a sequence of real numbers converges iff it is Cauchy. Thus, in your last inequality, you should write
$$
|x_n-x_m| \leq |S_n-S_m| \to 0 \text{ as }n,m\to \infty.
$$
A: Observe that for $\;n\,,\,m=n+k\;,\;\;k\in\Bbb N\;$ , we have
$$|x_m-x_n|=|x_{n+k}-x_n|=|x_{n+k}-x_{n+k-1}+x_{n+k-1}-x_{n+k-2}+\ldots+x_{n+1}-x_n|\le$$
$$\le|x_{n+k}-x_{n+k-1}|+\ldots+|x_{n+1}-x_n|\le\frac1{(n+k-1)^2}+\frac1{(x+k-2)^2}+\ldots+\frac1{n^2}\le$$
Now just observe that
$$\lim_{n\to\infty}\sum_{i=1}^n\frac1{(n+k-i)^2}=0\;,\;\;\text{since the series}\;\;\sum_{n=1}^\infty\frac1{n^2}$$
converges....fill up the (small, in fact) details left now.
A: Assume $N>n>2$. Then:
$$\begin{eqnarray*}\left|x_N-x_n\right|&\leq& |x_n-x_{n+1}|+|x_{n+1}-x_{n+2}|+\ldots+|x_{N-1}-x_N|\\&\leq&\frac{1}{n^2}+\frac{1}{(n+1)^2}+\ldots+\frac{1}{(N-1)^2}\\&\leq&\int_{n-1}^{+\infty}\frac{dx}{x^2}=\frac{1}{n-1}\end{eqnarray*}$$
so $\{x_n\}_{n\in\mathbb{N}}$ is a Cauchy sequence.
