real analysis - Find $limlimits_{ntoinfty}sumlimits_{i=1}^{n}{frac{2n}{(n+2i)^2}}$

Find $$\lim\limits_{n\to\infty}\sum\limits_{i=1}^{n}{\frac{2n}{(n+2i)^2}}.$$

I have tried dividing through by $1/n^2$ and various other algebraic tricks but cannot seem to make any progress on this limit. Wolfram Alpha gives the value as $2/3$, but I could not understand its derivation. Any insight is welcome.

Answer

The expression
$$\sum_{i = 1}^n \frac{2n}{(n+2i)^2} =\frac{1}{n} \sum_{i = 1}^n \frac{2}{(1+\frac{2i}{n})^2}$$
is the Riemann sum of the function $f(x)= \frac{2}{(1+2x)^2}$ over the interval $I = [0,1]$, corresponding to the uniform partition of $I$ into $n$ equal parts.

Since $f$ is Riemann-integrable (being a continuous function over a closed and bounded interval), this sum approaches the integral of $f$ over $I$ as $n \to \infty$. That is,
$$\begin{eqnarray*} \lim_{n \to \infty} \frac{1}{n} \sum_{i = 1}^n \frac{2}{(1+\frac{2i}{n})^2} &=& \int_0^1 \frac{2}{(1+2x)^2} \mathrm{d} \, x \\ &=& \left. -\frac{1}{1+2x} \right|_{x=0}^{x=1} \quad = \quad \frac{2}{3}. \end{eqnarray*}$$