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?

Your limit can be rewritten as a Riemann sum + $a_n$ where $a_n\rightarrow 0$.
This is a Riemann sum:
\begin{align}
\lim_{n\rightarrow \infty} \frac{1}{(n+1)^2}+ \frac{2}{(n+2)^2}+\dots+\frac{n}{(2n)^2}&=\lim_{n\rightarrow \infty} \frac{1}{n}\left ( \frac{1/n}{(1+1/n)^2}+ \frac{2/n}{(1+2/n)^2}+\dots+\frac{n/n}{(1+n/n)^2}\right)\\
&=\int_{0}^1 \frac{x}{(1+x)^2} \text{d}x \\
&=\int_{0}^1 \frac{x+1}{(1+x)^2} \text{d}x -\int_{0}^1 \frac{1}{(1+x)^2} \text{d}x\\
&=\int_{0}^1 \frac{1}{1+x} \text{d}x -\int_{0}^1 \frac{1}{(1+x)^2} \text{d}x
\end{align}
Now,
\begin{align}a_n&=\frac{1}{n^2} + \frac{1}{(n+1)^2}+ \frac{1}{(n+2)^2}+\dots+\frac{1}{(2n)^2} \\
&\leq \frac{1}{n^2} + \frac{1}{n^2}+ \frac{1}{n^2}+\dots+\frac{1}{n^2}\\
&=(n+1)\frac{1}{n^2}\rightarrow 0
\end{align}
