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.

$$
n\sum_{k=n}^\infty\frac1{k^2}=\sum_{k=n}^\infty\frac{n^2}{k^2}\frac1n\tag{1}
$$
is a Riemann sum ($x=\frac{k}{n}$ and $\mathrm{d}x=\frac1n$) for
$$
\int_1^\infty\frac1{x^2}\mathrm{d}x=1\tag{2}
$$
Although $(2)$ is an improper Riemann integral, because $\frac1{x^2}$ is decreasing, we can still use the rectangular upper and lower approximations to $(2)$:
$$
\sum_{k=n+1}^\infty\frac{n^2}{k^2}\frac1n<\int_1^\infty\frac1{x^2}\mathrm{d}x< \sum_{k=n}^\infty\frac{n^2}{k^2}\frac1n\tag{3}
$$
where the sums in $(3)$ differ by $\frac1n$. Therefore, combining $(1)$-$(3)$ yields
$$
1<n\sum_{k=n}^\infty\frac{1}{k^2}<1+\frac1n\tag{4}
$$
which by the sandwich theorem gives
$$
\lim_{n\to\infty}n\sum_{k=n}^\infty\frac{1}{k^2}=1\tag{5}
$$
