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edit approved Jan 30, 2017 at 18:49
Peter Mortensen
edit approved Jan 30, 2017 at 18:49
Peter Mortensen
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a A special binomial identity in need of a proof

Question. I've encountered a curious identity as a codicil in some work. Any proof/reference? $$\sum_{k=-n}^n\frac{2k+1}{n+k+1}\binom{2n}{n-k}\frac{x^k}{1+x^{2k+1}}=\frac{x^n}{1+x^{2n+1}}.$$

I've encountered a curious identity as a codicil in some work. Is there a proof or reference?

$$\sum_{k=-n}^n\frac{2k+1}{n+k+1}\binom{2n}{n-k}\frac{x^k}{1+x^{2k+1}}=\frac{x^n}{1+x^{2n+1}}.$$

a special binomial identity in need of a proof

Question. I've encountered a curious identity as a codicil in some work. Any proof/reference? $$\sum_{k=-n}^n\frac{2k+1}{n+k+1}\binom{2n}{n-k}\frac{x^k}{1+x^{2k+1}}=\frac{x^n}{1+x^{2n+1}}.$$

A special binomial identity in need of a proof

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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T. Amdeberhan
T. Amdeberhan
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a special binomial identity in need of a proof

Question. I've encountered a curious identity as a codicil in some work. Any proof/reference? $$\sum_{k=-n}^n\frac{2k+1}{n+k+1}\binom{2n}{n-k}\frac{x^k}{1+x^{2k+1}}=\frac{x^n}{1+x^{2n+1}}.$$