find limit of $a_n=\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+...+\frac{1}{(2n)^2}$ finding limit of $a_n=\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+...+\frac{1}{(2n)^2}$ 
I know that i have to use Stolz Cesaro theorem, but the problem is that i need second sequence.

$$a_n=\frac{1}{n^2}\left( \frac{1}{(1+\frac1n)^2}+\frac{1}{(1+\frac2n)^2}+...+\frac{1}{(1+\frac{n}n)^2} \right)=\frac{1}{n}\left[\frac{1}{n}\left( \frac{1}{(1+\frac1n)^2}+\frac{1}{(1+\frac2n)^2}+...+\frac{1}{(1+\frac{n}n)^2} \right)\right]$$
Now, apply  S-C to
$$a_n=\frac{ \left( \frac{1}{(1+\frac1n)^2}+\frac{1}{(1+\frac2n)^2}+...+\frac{1}{(1+\frac{n}n)^2} \right)  }{n^2}$$
But this is an overkill, it is easy to see that the top is $\leq 1+1+1..+1=n$.
P.S.
$$\frac{1}{n}\left( \frac{1}{(1+\frac1n)^2}+\frac{1}{(1+\frac2n)^2}+...+\frac{1}{(1+\frac{n}n)^2} \right)$$ 
is a Riemann sum, thus convergent, and $\frac{1}{n}$ converges to $0$. 
Alternately, if you don't know integrals,apply S-C to
$$a_n=\frac{ \left( \frac{1}{(1+\frac1n)^2}+\frac{1}{(1+\frac2n)^2}+...+\frac{1}{(1+\frac{n}n)^2} \right)  }{n^2}$$
