Sum of an infinite series of fractions involving multiple terms in the denominator

Question

This is the series in question:

$$S = \frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\frac{3}{1+3^2+3^4} \ldots$$

The general term seems to be:

$$T_n=\frac{n}{1+n^2+n^4}$$

In the original question, which is from Resonance DLPD Algebra, the value of $14S$ is asked.

I haven't been able to make any progress on this question even after trying for a long time. I don't know how to handle series where the denominator contains sum of numbers in general. Can you please provide a hint?

Notice that $14 = 1^2 + 2^2 + 3^2$ I doubt that is a coincidence :) — Zubin Mukerjee, Feb 05 '19 at 04:34
https://math.stackexchange.com/questions/449510/how-to-find-the-sum-of-the-sequence-frac111214-frac212224 — lab bhattacharjee, Feb 05 '19 at 04:35
See also https://math.stackexchange.com/questions/304851/evaluate-sum-limits-k-1-infty-frack2-1k4k21/304858#304858 — lab bhattacharjee, Feb 05 '19 at 04:36

score 1 · Accepted Answer · answered Feb 05 '19 at 06:12

We are looking for: $S_n=\sum\limits_{n=1}^\infty\frac{n}{1+n^2+n^4}$

Let's realize that $\frac{n}{1+n^2+n^4}=\frac{1}{2}\big(\frac{1}{1-n+n^2}-\frac{1}{1+n+n^2}\Big)=\frac{1}{2}\big(\frac{1}{(n-\frac{1}{2})^2+\frac{3}{4}}-\frac{1}{(n+\frac{1}{2})^2+\frac{3}{4}}\Big)$

Which provide telescopic sum so $S=\frac{1}{2}$

Sum of an infinite series of fractions involving multiple terms in the denominator

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