maybe done before

maybe this has been doen before..but i ma not getting link...

if it hasnt been..then plz help

_n=1Î£^âˆžn4n⁴+1

2 Answers

106

Asish Mahapatra ·2010-01-26 18:25:36

\sum_{n=1}^{\infty }{\frac{n}{4n^4+1}}
=\sum_{n=1}^{\infty }{\frac{n}{4n^4+4n^2+1-4n^2}}

=\sum_{n=1}^{\infty }{\frac{n}{(2n^2+1)^2-(2n)^2}}

=\frac{1}{4}\sum_{n=1}^{\infty }{\frac{(2n^2+2n+1)-(2n^2-2n+1)}{(2n^2+2n+1)(2n^2-2n+1)}}

now i think it can be done

btw eure i had asked exactly this ques to u long back and u had solved then (i think at gsgj)

thx dude...

and i keep forgetting things...thats one problem[2]

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