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Problem :

How to find the sum of the sequence $\frac{1}{1+1^2+1^4} +\frac{2}{1+2^2+2^4} +\frac{3}{1+3^2+3^4}+.....$

I am unable to find out how to proceed in this problem.. this is a problem of arithmetic progression... Please suggest how to proceed...Thanks..

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  • $\begingroup$ I just did this question today, seemed rather familiar when I saw it, I would be interested to know the source where you got this problem :) $\endgroup$ – WhizKid Jul 22 '13 at 16:32
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HINT:

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

and $(1+r+r^2)-(1-r+r^2)=2r,$

$$\frac r{1+r^2+r^4}$$

$$=\frac12\cdot\frac{2r}{(1+r+r^2)(1-r+r^2)}$$

$$=\frac12\cdot\frac{(1+r+r^2)-(1-r+r^2)}{(1+r+r^2)(1-r+r^2)}$$

$$=\frac12\left(\frac1{1-r+r^2}-\frac1{1+r+r^2}\right)$$

Put the values of $r=1,2,\cdots.. n-1,n$ to find the partial sum and recognize the Telescoping Sum/Series which is evident as $1-(r+1)+(r+1)^2=1+r+r^2$

Then set $n\to\infty$

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