Find $ \lim_{n\rightarrow \infty} \frac{1}{n^2} + \frac{2}{(n+1)^2}+\dots+\frac{n+1}{(2n)^2} $ I want to find the following limit: 
$$ \lim_{n\rightarrow \infty} \frac{1}{n^2} + \frac{2}{(n+1)^2}+ \frac{3}{(n+2)^2}+\dots+\frac{n+1}{(2n)^2} $$
I guess it converges to 0 and I tried to prove it using the squeeze theorem but it doesn't seem to work.
Any ideas?

Note that
$$\frac{1}{n^2} + \frac{2}{(n+1)^2}+ \frac{3}{(n+2)^2}+\dots+\frac{n+1}{(2n)^2}=\sum_{k=n}^{2n}\frac{k-n+1}{k^2}
=\sum_{k=n}^{2n}\frac{1}{k}-(n-1)\sum_{k=n}^{2n}\frac{1}{k^2}\sim \log \left(\frac{2n}{n-1}\right)-\frac{n+1}{2n}\to\log 2-\frac12$$
indeed for Harmonic series and Euler–Maclaurin formula
$$\sum_{k=n}^{2n}\frac{1}{k}=\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^{n-1}\frac{1}{k}\sim\log2n-\log (2n-1)=\log \left(\frac{2n}{n-1}\right)$$
$$\sum_{k=n}^{2n}\frac{1}{k^2}=\sum_{k=1}^{2n}\frac{1}{k^2}-\sum_{k=1}^{n-1}\frac{1}{k^2}\sim -\frac1{2n}+\frac{1}{n-1}=\frac{n+1}{2n(n-1)}$$
