Evaluate $\lim_{n\rightarrow \infty}\sum_{r=1}^{n}\frac{n}{(n+r)^2}$ $$
\lim_{n\ \to\ \infty}\left[\,%
{n \over \left(\, n + 1\,\right)^{2}}+
{n \over \left(\, n + 2\,\right)^{2}}+\cdots +
{n \over \left(\, 2n\,\right)^{2}}\,\right]
$$
How can I deal with this limit in a reasonable way ?.
EDIT: I am STILL looking for a solution without use of integration 
 A: 
I am looking for a solution without use of calculus

I'm interpreting that as "without integration", since limits are calculus.
Write
$$\begin{aligned}
\frac{1}{(n+r)^2} &= \frac{1}{(n+r)(n+r+1)} + \left(\frac{1}{(n+r)^2}-\frac{1}{(n+r)(n+r+1)}\right)\\
&= \left(\frac{1}{n+r} - \frac{1}{n+r+1}\right) + \frac{1}{(n+1)^2(n+r+1)}.
\end{aligned}$$
Then you get
\begin{align}
\sum_{r=1}^n \frac{n}{(n+r)^2} &= n\sum_{r=1}^n \left(\frac{1}{n+r}-\frac{1}{n+r+1}\right) + \underbrace{\sum_{r=1}^n \frac{n}{(n+r)^2(n+r+1)}}_{R_n}\\
&= \underbrace{n\left(\frac{1}{n+1} - \frac{1}{n+n+1}\right)}_{C_n} + R_n.
\end{align}
It is easily seen that $C_n \to \frac{1}{2}$, and $R_n\to 0$ follows since it is the sum of $n$ positive terms each of which is smaller than $\frac{1}{n^2}$, so $0 < R_n < \frac{1}{n}$.
A: $$\sum_{r=1}^n\frac n{(n+r)^2}=\frac1n\sum_{r=1}^n\frac{n^2}{(n+r)^2}=\frac1n\sum_{r=1}^n\frac1{(1+r/n)^2}$$
Now see Find $\lim\limits_{n \to \infty} \frac{1}{n}\sum\limits^{2n}_{r =1} \frac{r}{\sqrt{n^2+r^2}}$
