Let $x_n$ be a sequence in a complete metric space $(X,d)$ such that $d(x_n,x_{n+1}) \le \frac {1}{n^2}$. Is $x_n$ convergent? I have a proof that $x_n$ is convergent. Here it is:
Let $\epsilon \gt 0$ be given and let $m \gt n$ where $n,m \in \Bbb N$.
Then $d(x_n,x_m)\le d(x_n,x_{n+1})+d(x_{n+1},x_{n+2})+...+d(x_{m-1},x_m) \le \frac 1{n^2} +\frac 1{(n+1)^2}+...\frac 1{(m-1)^2}=\sum_{k=1}^{m-1}\frac 1{k^2}-\sum_{k=1}^{n-1} \frac {1}{k^2}.$
But since we know that $\sum \frac 1{n^2}$ is convergent. hence it's sequence of partial sums is Cauchy. So there exists $N \in \Bbb N$ such that $m-1, n-1 \ge N \Rightarrow  \sum_{k=1}^{m-1} \frac 1{k^2} - \sum_{k=1}^{n-1} \frac {1}{k^2} \lt \epsilon$.
Hence $d(x_n,x_m) \le \epsilon$ whenever $m,n \gt N$. So $x_n$ is a Cauchy sequence in a complete metric space hence it is convergent.
QED.

My question is that is it possible to show that $x_n$ is Cauchy without using convergence of $\sum \frac 1{n^2}$?
I mean from "$d(x_n,x_m) \le \frac 1{n^2} +\frac 1{(n+1)^2}+...\frac 1{(m-1)^2}$", what manipulations can be done without introducing $\sum \frac 1{n^2}$?
For example : In case of the expression $\frac 1{n!} + \frac 1{(n+1)!} + ... + \frac 1{(m-1)!}$, we can use the inequality $2^n \lt n! (\text {for} \; n \gt 3)$ such that $\frac 1{n!} + \frac 1{(n+1)!} + ... + \frac 1{(m-1)!} \lt \frac 1{2^n}+\frac 1{2^{n+1}}+...+ \frac 1{2^{m-1}} \lt \frac 1{2^{n-1}}$.
Then we can choose $n \gt 1-\log_2\epsilon$ so that $\frac 1{2^{n-1}} \lt \epsilon$. 
I tried this : $\frac 1{n^2} +\frac 1{(n+1)^2}+...\frac 1{(m-1)^2} \lt \frac {m-n}{n^2}.$ But this $m-n$ in numerator is bugging me.

How about 
\begin{align*}
d(x_n,x_m) \le \frac 1{n^2} +\frac 1{(n+1)^2}+...\frac 1{(m-1)^2}&\le \frac 1{n(n-1)} +\frac 1{(n+1)n}+...\frac 1{(m-1)(m-2)}\\
&\le \frac 1{n-1} -\frac 1{m-1}\le\frac 1{n-1},
\end{align*}
but this is in fact one of the approaches to show $\sum_{n=1}^\infty \frac{1}{n^2}$ is convergent.
