Find the sum to `n` terms of the series: `1/(1+1^2+1^4)+2/(1+2^2+2^4)+3/(1+3^2+3^4)+`

Explanation

Given,`1/(1+1^2+1^4)+2/(1+2^2+2^4)+3/(1+3^2+3^4)+....`

The `m^(th)` term of the series is,

`m/(1+m^2+m^4)=m/((1+m^2)^2-m^2)=m/[(1+m+m^2)(1-m+m^2)]`

=`1/[2.(1-m+m^2)]-1/[2.(1+m+m^2)]`

The positive part of `(m+1)^(th)` term is same as the negative part of the `m^(th)` term.

so adding we get,`(m^2+m)/(2(1+m^2+m))`