Question

Find the sum to $n$ terms of the series:$\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\frac{3}{1+3^2+3^4}+.........$

Answer 1

Given in an series

$\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\frac{3}{1+3^2+3^4}+.......$

The general term of the above series is $T_n=\frac{n}{1+n^2+n^4}$

We can rearrange the terms of the general term to simplify it and write as

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

Now we can split the above expression into two constituting term such that the sum of the $n$ terms of the series can be found out easily.

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

Now, for

$n=1T_1=\frac{1}{2}[1-\frac{1}{3}]=\frac{1}{2}-\frac{1}{6}n=2T_2=\frac{1}{2}[\frac{1}{3}-\frac{1}{7}]=\frac{1}{6}-\frac{1}{14}n=3T_3=\frac{1}{2}[\frac{1}{7}-\frac{1}{13}]=\frac{1}{14}-\frac{1}{26}....\therefore S={\sum }_{i=1}^{n}i{T}_1+T_2+T_3+.......T_n=\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{14}+\frac{1}{14}.......-\frac{1}{2}\left[\frac{1}{n^2+1-n}\right]+\frac{1}{2}\left[\frac{1}{n^2+1-n}\right]-\frac{1}{2}\left[\frac{1}{n^2+1+n}\right]\therefore S=\frac{1}{2}-\frac{1}{2}\left[\frac{1}{n^2+1-n}\right]=\frac{n^2+n}{2(n^2+n+1)}$