Double-Square Summation

Note first that

$\dfrac{n}{n^{4} + 4} = \dfrac{1}{4}\left(\dfrac{1}{n^{2} - 2n + 2} - \dfrac{1}{n^{2} + 2n + 2}\right) =$

$\dfrac{1}{4}\left(\dfrac{1}{(n - 1)^{2} + 1} - \dfrac{1}{(n + 1)^{2} + 1}\right).$

When this is summed over the positive integers, we end up with a telescoping sum, with terms canceling pairwise and leaving us with only the first "parts" of the first two terms, i.e., we end up with

$\dfrac{1}{4}\left(\dfrac{1}{(1 - 1)^{2} + 1}\right) + \dfrac{1}{4}\left(\dfrac{1}{(2 - 1)^{2} + 1}\right) = \dfrac{1}{4} + \dfrac{1}{8} = \dfrac{3}{8} = \boxed{0.375}.$

$\dfrac n {n^4 + 4} = \dfrac 1 4 \left ( \dfrac1 {n^2 - 2n + 2} - \dfrac 1 {n^2 + 2n + 2} \right) \ \ \ \ \ \ By\ partial\ fractions.\\ \therefore\ 4*\dfrac n {n^4 + 4} \\ =\displaystyle \sum_1^{\infty} \dfrac1 {n^{2} - 2n + 2} - \sum_1^{\infty} \dfrac1 {n^{2} + 2n + 2}\\ =\displaystyle \sum_1^2 \dfrac1 {n^{2} - 2n + 2} + \sum_3^{\infty} \dfrac1 {n^{2} - 2n + 2}- \sum_1^{\infty} \dfrac1 {n^{2} + 2n + 2} \\ =\displaystyle 1+\frac 1 2 + \sum_1^{\infty} \dfrac1 {n^{2} + 2n + 2}- \sum_1^{\infty} \dfrac1 {n^{2} + 2n + 2} \\ =\frac 3 2.\\ \therefore\ \displaystyle \sum_1^{\infty} \dfrac n {n^4 + 4}=\frac 3 8.$

Upon solving the last step shows that substituting n with [n+2] we get the second term... So cancellation of terms begins every two times