Question

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

• A. $\frac{1}{4}$
• B. $\frac{1}{3}$
• C. $\frac{3}{8}$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $\frac{3}{8}$

$T_n=\frac{n}{n^4+4}{T}_n=\frac{n}{n^4+4n^2+4-4n^2}=\frac{n}{{(n^2+2)}^2-{\left(2n\right)}^2}=\frac{n}{(n^2+2n+2)(n^2-2n+2)}=\frac{n\left[(n^2+2n+2)(n^2-2n+2)\right]}{(n^2+2n+2)(n^2-2n+2)}\times \frac{1}{4n}=\frac{1}{4}[\frac{1}{n^2-2n+2}-\frac{1}{n^2+2n+2}]T_1=\frac{1}{4}[\frac{1}{1}-\frac{1}{5}]$

$T_2=\frac{1}{4}[\frac{1}{2}-\frac{1}{10}]T_3=\frac{1}{4}[\frac{1}{5}-\frac{1}{17}]T_4=\frac{1}{4}[\frac{1}{10}-\frac{1}{26}]...{\sum }_{n=1}^{\infty }\frac{1}{4}[1+\frac{1}{2}]=\frac{3}{8}$