proving convergence of a sequence and then finding its limit For every $n$ in $\mathbb{N}$, let: $$a_{n}=n\sum_{k=n}^{\infty }\frac{1}{k^{2}}$$
Show that the sequence $\left \{ a_{n} \right \}$ is convergent and then calculate its limit.
To prove it is convergent, I was thinking of using theorems like the monotone convergence theorem. Obviously, all the terms $a_{n}$ are positive. So, if I prove that the sequence is decreasing, then by the monotone convergence theorem it follows that the sequence itself is convergent. $a_{n+1}-a_{n}=-\frac{1}{n}+\sum_{k=n+1}^{\infty }\frac{1}{k^{2}}$. But, I can't tell from this that the difference $a_{n+1}-a_{n}$ is negative. 
If anybody knows how to solve this problem, please share.

[Edit: the first proof or convergence wasn't quite right, so I removed it.]
It is useful to find some estimates first (valid for $n>1$):
$$\sum_{k=n}^{\infty }\frac{1}{k^{2}}<\sum_{k=n}^{\infty }\frac{1}{k(k-1)}=\sum_{k=n}^{\infty }\left(\frac1{k-1}-\frac1k\right)=\frac1{n-1}\\\sum_{k=n}^{\infty }\frac{1}{k^{2}}>\sum_{k=n}^{\infty }\frac{1}{k(k+1)}=\sum_{k=n}^{\infty }\left(\frac1{k}-\frac1{k+1}\right)=\frac1{n}$$
The last equality in each of these lines holds because those are telescoping series.
This gives us the estimate: $1<a_n<\frac{n}{n-1}$. By the squeeze theorem, we can conclude that our sequence converges and $\lim_{n\to\infty}a_n=1$.
