If $S=\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\cdots+\frac{n}{1+n^2+n^4}$, then calculate $14S$.

Question

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

The question can be simplified to:

Find $S=\sum\limits_{k=1}^n\,t_k$ if $t_n=\dfrac{n}{1+n^2+n^4}$.

As $1+n^2+n^4$ forms a GP,
$$t_n=\frac{n(n^2-1)}{n^6-1}\,.$$

But I can't figure out how to solve further. It would be great if someone could help.

Answer 1

HINT:

As $n^4+n^2+1=(n^2+1)^2-n^2=(n^2-n+1)(n^2+n+1)$

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

If $f(m)=\dfrac1{m^2-m+1}, f(m+1)=?$

See Telescoping series

Answer 2

From the first four terms, you can easily guess the pattern: The numerators are $1,3,6,10$ which look like $n(n+1)/2$ and the denominators are $3,7,13,21$ which look to have a constant second difference of $2$ and are thus probably $n^2+n+1$. This motivates trying to prove, by induction, that $$ S_n = \frac{n(n+1)}{2(n^2+n+1)} $$ The basis is trivial to establish for $n=1$ and the induction is fairly simple algebra.