Tuesday, March 1, 2016

real analysis - Compute: suminftyk=1suminftyn=1frac1k2n+2nk+n2k



I try to solve the following sum:


k=1n=11k2n+2nk+n2k


I'm very curious about the possible approaching ways that lead us to solve it. I'm not experienced with these sums, and any hint, suggestion is very welcome. Thanks.


Answer



I think one way of approaching this sum would be to use the partial fraction 1k2n+2nk+n2k=1kn(k+n+2)=12(1k+1n)(1k+n1n+k+2) to rewrite you sum in the form n=1k=11k2n+n2k+2kn=12k=1n=1(1n(k+n)1n(k+n+2)+1k(k+n)1k(k+n+2)) Since the sum on the right will telescope in one of the summation variables it should be straightforward to find the answer from here (it ends up being 7/4 I think).


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