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Find the sum of the $n$ terms of the following series: $$\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\frac{3}{1+3^2+3^4}+\dots$$ [Original Image]

I tried it a lot .

But don't get a proper start .

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    First, derive this relation: $\sum_{n=0}^m \frac{n}{1+n^2+n^4} = \frac{m^2+m}{2(m^2+m+1)}$. You can try this by induction, but look for a neat method if possible.Take $m \to \infty$ to see what happens, – 2017-01-26
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    @астонвіллаолофмэллбэрг It's easy to prove that relation using induction but how did you find it in the first place? – 2017-01-26
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    @Hugh do PFD and telescoping. – 2017-01-26
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    @SimplyBeautifulArt Simply Beautifully said. – 2017-01-26

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The $k$th term of the series is $$\frac {k}{1+k^2+k^4} = \frac {k}{(1+k^2)^2-k^2} = \frac {k}{(1+k^2+k)(1+k^2-k)} = \frac {1}{2 (1+k^2-k)} - \frac {1}{2 (1+k^2+k)} $$ Thus, we have a telescoping series. Thus, the sum of the first $n $ terms is: $$\frac {1}{2 (1+1^2-1)} - \frac {1}{2(1+n^2+n)} =\frac {n^2+n}{2 (1+n+n^2)} $$ Hope it helps.

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    :There is a typo in 2nd line 2nd term. – 2017-01-26
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    @MatheMagic Thank you. I was on my phone. Thank you again. – 2017-01-26