I've encountered a curious identity as a codicil in some work. Is there a proof or reference?
$$\sum_{k=-n}^n\frac{2k+1}{n+k+1}\binom{2n}{n-k}\frac{x^k}{1+x^{2k+1}}=\frac{x^n}{1+x^{2n+1}}.$$
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asked Jan 30, 2017 at 2:57
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1$\begingroup$ The right-hand denominator can be found here: en.wikipedia.org/wiki/Betti_number $\endgroup$– Fred Daniel KlineCommented Jan 30, 2017 at 3:32
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1 Answer
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The $k=j-1$ and $k=-j$ terms cancel, so all that's left is the $k=n$ term.
