Question

The sum of infinity terms of the series $\frac{1}{1+1^2+1^4}+\frac{1}{1+2^2+2^4}+\frac{3}{1+3^2+3^4}+....\infty$ is

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

Answer 1

The correct option is A $\frac{1}{2}$

$\begin{array}{cc}{T}_n& =\frac{n}{1+n^2+n^4}\\ & =\frac{n}{{(1+n^2)}^2-n^2}\\ & =\frac{n}{(1+n^2+n)(1+n^2-n)}\\ & =\frac{1}{2}\left\{\frac{(1+n^2+n)-(1+n^2-n)}{(1+n^2+n)(1+n^2-n)}\right\}\end{array}$

$T_n=\frac{1}{2}\{\frac{1}{1+n^2-n}-\frac{1}{1+n^2+n}\}$

$\begin{array}{cc}\sum _{n=1}^{\infty }{T}_n=& \frac{1}{2}(\frac{1}{1}-\frac{1}{3})\\ & +\frac{1}{2}(\frac{1}{3}-\frac{1}{7})\\ & +\frac{1}{2}(\frac{1}{7}-\frac{1}{13})\\ & =\frac{1}{2}\left(1\right)\end{array}$

Answer Option (A).