# Find $$\displaystyle\lim_{{{n}\to\infty}}{\sum_{{{i}={1}}}^{{{n}}}}{\frac{{{2}{n}}}{{{\left({n}+{2}{i}\right)}^{{2}}}}}$$

Find
$\underset{n\to \mathrm{\infty }}{lim}\sum _{i=1}^{n}\frac{2n}{{\left(n+2i\right)}^{2}}$
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Drahthaare89c
The expression
$\sum _{i=1}^{n}\frac{2n}{{\left(n+2i\right)}^{2}}=\frac{1}{n}\sum _{i=1}^{n}\frac{2}{{\left(1+\frac{2i}{n}\right)}^{2}}$
is the Riemann sum of the function $f\left(x\right)=\frac{2}{{\left(1+2x\right)}^{2}}$ over the interval $I=\left[0,1\right]$, 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 $n\to \mathrm{\infty }$. That is,
$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}\sum _{i=1}^{n}\frac{2}{{\left(1+\frac{2i}{n}\right)}^{2}}={\int }_{0}^{1}\frac{2}{{\left(1+2x\right)}^{2}}dx$
$=-\frac{1}{1+2x}{\mid }_{x=0}^{x=1}=\frac{2}{3}$
###### Not exactly what you’re looking for?
Alannah Campos
You can also use Euler-Maclaurin summation.
The first-order Euler-Maclaurin formula says
$\sum _{i=1}^{n}f\left(i\right)={\int }_{1}^{n}f\left(x\right),dx+\left\{f\left(1\right)+f\left(n\right)2+{\int }_{1}^{n}{f}^{\prime }\left(x\right)\left(x-⌊x⌋-\frac{1}{2}\right),dx.$
Since $|x-⌊x⌋-\frac{1}{2}|\le 1$, with $f\left(x\right)=\frac{2n}{{\left(n+2x\right)}^{2}}$ we have
$\sum _{i=1}^{n}\frac{2n}{{\left(n+2i\right)}^{2}}={\int }_{1}^{n}\frac{2n,dx}{{\left(n+2x\right)}^{2}}+{R}_{n},$
where $|{R}_{n}|\le |\frac{3f\left(n\right)-f\left(1\right)}{2}|=|\frac{3}{n}-\frac{n}{{\left(n+2\right)}^{2}}|$
Therefore,
$\underset{n\to \mathrm{\infty }}{lim}\sum _{i=1}^{n}\frac{2n}{{\left(n+2i\right)}^{2}}=\underset{n\to \mathrm{\infty }}{lim}{\int }_{1}^{n}\frac{2n,dx}{{\left(n+2x\right)}^{2}}$
$=\underset{n\to \mathrm{\infty }}{lim}{\left[\frac{-n}{n+2x}\right]}_{1}^{n}$
$=\underset{n\to \mathrm{\infty }}{lim}\left(-\frac{1}{3}+\frac{n}{n+2}\right)$
$=\frac{2}{3}$