Find $ \lim\limits_{n\to\infty}\sum\limits_{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.
 A: 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,
$$
\begin{eqnarray*}
\lim_{n \to \infty} \frac{1}{n} \sum_{i = 1}^n \frac{2}{(1+\frac{2i}{n})^2} 
&=& 
\int_0^1 \frac{2}{(1+2x)^2} \mathrm{d} \, x
\\ &=&
\left. -\frac{1}{1+2x} \right|_{x=0}^{x=1}
\quad = \quad
\frac{2}{3}.
\end{eqnarray*}
$$
A: You can also use Euler-Maclaurin summation.
The first-order Euler-Maclaurin formula says $$\sum_{i=1}^n f(i) = \int_1^n f(x) \, dx + {f(1) + f(n) \over 2} + \int_1^n f'(x) \left(x - \lfloor x \rfloor - \frac{1}{2}\right)\,dx.$$
Since $|x - \lfloor x \rfloor - \frac{1}{2}| \leq 1$, with $f(x) = \frac{2n}{(n+2x)^2}$ 
we have
$$\sum_{i=1}^n \frac{2n}{(n+2i)^2} = \int_1^n \frac{2n \, dx}{(n+2x)^2} +  R_n,$$
where $|R_n| \leq \left|\frac{3f(n) - f(1)}{2}\right| = \left|\frac{3}{n} - \frac{n}{(n+2)^2}\right|$. 
Therefore, 
$$\lim_{n \to \infty} \sum_{i=1}^n \frac{2n}{(n+2i)^2} = \lim_{n \to \infty} \int_1^n \frac{2n \, dx}{(n+2x)^2} = \lim_{n \to \infty} \left[\frac{-n }{n+2x} \right]_1^n = \lim_{n \to \infty} \left(-\frac{1}{3} +  \frac{n}{n+2}\right) = \frac{2}{3}.$$
