Find

$\underset{n\to \mathrm{\infty}}{lim}\sum _{i=1}^{n}\frac{2n}{{(n+2i)}^{2}}$

palmantkf4u
2022-04-05
Answered

Find

$\underset{n\to \mathrm{\infty}}{lim}\sum _{i=1}^{n}\frac{2n}{{(n+2i)}^{2}}$

You can still ask an expert for help

Drahthaare89c

Answered 2022-04-06
Author has **19** answers

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$f\left(x\right)=\frac{2}{{(1+2x)}^{2}}$ 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 $n\to \mathrm{\infty}$ . That is,

$\underset{n\to \mathrm{\infty}}{lim}\frac{1}{n}\sum _{i=1}^{n}\frac{2}{{(1+\frac{2i}{n})}^{2}}={\int}_{0}^{1}\frac{2}{{(1+2x)}^{2}}dx$

$=-\frac{1}{1+2x}{\mid}_{x=0}^{x=1}=\frac{2}{3}$

is the Riemann sum of the function

Alannah Campos

Answered 2022-04-07
Author has **10** answers

You can also use Euler-Maclaurin summation.

The first-order Euler-Maclaurin formula says

$\sum _{i=1}^{n}f\left(i\right)={\int}_{1}^{n}f\left(x\right),dx+\{f\left(1\right)+f\left(n\right)2+{\int}_{1}^{n}{f}^{\prime}\left(x\right)(x-\lfloor x\rfloor -\frac{1}{2}),dx.$

Since$|x-\lfloor x\rfloor -\frac{1}{2}|\le 1$ , with $f\left(x\right)=\frac{2n}{{(n+2x)}^{2}}$ we have

$\sum _{i=1}^{n}\frac{2n}{{(n+2i)}^{2}}={\int}_{1}^{n}\frac{2n,dx}{{(n+2x)}^{2}}+{R}_{n},$

where$\left|{R}_{n}\right|\le \left|\frac{3f\left(n\right)-f\left(1\right)}{2}\right|=|\frac{3}{n}-\frac{n}{{(n+2)}^{2}}|$

Therefore,

$\underset{n\to \mathrm{\infty}}{lim}\sum _{i=1}^{n}\frac{2n}{{(n+2i)}^{2}}=\underset{n\to \mathrm{\infty}}{lim}{\int}_{1}^{n}\frac{2n,dx}{{(n+2x)}^{2}}$

$=\underset{n\to \mathrm{\infty}}{lim}{\left[\frac{-n}{n+2x}\right]}_{1}^{n}$

$=\underset{n\to \mathrm{\infty}}{lim}(-\frac{1}{3}+\frac{n}{n+2})$

$=\frac{2}{3}$

The first-order Euler-Maclaurin formula says

Since

where

Therefore,

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