jmc

First, we can factor the denominator with a little give and take: $$\begin{array}{rl}{n}^4+4& =n^4+4n^2+4-4n^2\\ & =(n^2+2{)}^2-(2n{)}^2\\ & =(n^2+2n+2)(n^2-2n+2).\end{array}$$Then $$\begin{array}{rl}\sum _{n=1}^{\infty }\frac{n}{n^4+4}& =\sum _{n=1}^{\infty }\frac{n}{(n^2+2n+2)(n^2-2n+2)}\\ & =\frac{1}{4}\sum _{n=1}^{\infty }\frac{(n^2+2n+2)-(n^2-2n+2)}{(n^2+2n+2)(n^2-2n+2)}\\ & =\frac{1}{4}\sum _{n=1}^{\infty }\left(\frac{1}{n^2-2n+2}-\frac{1}{n^2+2n+2}\right)\\ & =\frac{1}{4}\sum _{n=1}^{\infty }\left(\frac{1}{(n-1{)}^2+1}-\frac{1}{(n+1{)}^2+1}\right)\\ & =\frac{1}{4}\left[\left(\frac{1}{0^2+1}-\frac{1}{2^2+1}\right)+\left(\frac{1}{1^2+1}-\frac{1}{3^2+1}\right)+\left(\frac{1}{2^2+1}-\frac{1}{4^2+1}\right)+\cdots \right].\end{array}$$Observe that the sum telescopes. From this we find that the answer is $\frac{1}{4}\left(\frac{1}{0^2+1}+\frac{1}{1^2+1}\right)=\boxed{\frac{3}{8}}$.