Question

If $S=\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\frac{3}{1+3^2+3^4}.......\infty$, then find the value of $14S.$

Answer 1

$S=\frac{1}{1{+1}^2+1^4}+\frac{2}{1+2^2+2^4}+...\infty S={\sum }_{n=1}^{\infty }\frac{n}{{1+n}^2+n^4}S=\frac{1}{2}{\sum }_{n=1}^{\infty }\frac{2}{{(n}^2+n+1){(n}^2-n+1)}S=\frac{1}{2}{\sum }_{n=1}^{\infty }(\frac{1}{{(n}^2-n+1)}-\frac{1}{{(n}^2+n+1)})S=\frac{1}{2}[(\frac{1}{1}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{7})+(\frac{1}{7}-\frac{1}{13})+....\infty ]S=\frac{1}{2}14S=7$