I try to solve the following sum:
∞∑k=1∞∑n=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+n−1n+k+2) to rewrite you sum in the form ∞∑n=1∞∑k=11k2n+n2k+2kn=12∞∑k=1∞∑n=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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