Question

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

• A. 1/4
• B. 1/3
• C. 3/8
• D. 1/2

Answer 1

Answer: (C)

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