Using properties of limits, calculate $\lim_{n\to \infty}\left(\frac1{n^2}+\frac1{(n+1)^2}+\cdots+\frac1{(2n)^2}\right)$ Using the properties of limits, calculate the following limits, if they exist. If not, prove they do not exist:
$$\lim_{n\to \infty}\left(\frac1{n^2}+\frac1{(n+1)^2}+\frac1{(n+2)^2}+\cdots+\frac1{(2n)^2}\right)$$
This is what I have done, I have expressed the limit in the form:
$\lim_{n\to \infty}\frac1{(n+a)^2}$ where 'a' belongs to the reals. 
Then using the $\epsilon-N$ definition of limits, I assumed that:
$$\lim_{n\to \infty}\frac1{(n+a)^2}=0$$ and carried forward with the proof. I would like to use the $\epsilon-N$ definition of limits since it is what we are covering right now, is this the right way of solving this problem?

$\sum_{n\in N}1/n^2$ converges. Therefore $$\lim_{m\to \infty}\sup_{n\geq m} \sum_{j=n}^{j =2 n}(1/n^2)\leq \lim_{m\to \infty}\sup_{n' \geq n\geq m}\sum_{j=n}^{j=n'}(1/n^2)=0.$$
In general if $\sum_{n\in N} a_n$ converges then $\lim_{n\to \infty}\sum _{j=n}^{j=2 n}a_j=0.$
The most elementary way to prove that $\sum_n (1/n^2)$ converges is that for $n>1$ we have $1/n^2<1/n(n-1)=1/(n-1)-1/n$, so for $n>1$ we have
 $$\sum_{j=1}^n(1/n^2)=1+\sum_{j=2}^n(1/n^2)<1+\sum_{j=2}^n[1/(j-1)-1/j]=2-1/n.$$
