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  4. The sum displaystyle∑n=1∞((n/n4+4)) is equal to

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Q. The sum $\displaystyle\sum_{n=1}^{\infty}\left(\frac{n}{n^4+4}\right)$ is equal to

Sequences and Series

Solution:

$T_n=\frac{n}{\left(n^4+4 n^2+4\right)-4 n^2}=\frac{n}{\left(n^2+2\right)^2-(2 n)^2}=\frac{n}{\left(n^2+2+2 n\right)\left(n^2+2-2 n\right)}$
$T_n=\frac{1}{4}\left[\frac{\left(n^2+2+2 n\right)-\left(n^2-2 n+2\right)}{\left(n^2+2+2 n\right)\left(n^2-2 n+2\right)}\right]=\frac{1}{4}\left[\frac{1}{(n-1)^2+1}-\frac{1}{(n+1)^2+1}\right]$
$S _{ n }=\displaystyle\sum_{ n =1}^{\infty} T _{ n }=\frac{3}{8}$