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T. Amdeberhan
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QUESTION. Let $x>0$ be a real number or an indeterminate. Is this true? $$\sum_{n=0}^{\infty}\frac{\binom{2n+1}{n+1}}{2^{2n+1}\,(n+x+1)}=\frac{2^{2x}}{x\,\binom{2x}x}-\frac1x.$$

POSTSCRIPT. I like to record this presentable form by Alexander Burstein: $$\sum_{n=0}^{\infty}\frac{\binom{2n}n}{2^{2n}(n+x)}=\frac{2^{2x}}{x\binom{2x}x}.$$

QUESTION. Let $x>0$ be a real number or an indeterminate. Is this true? $$\sum_{n=0}^{\infty}\frac{\binom{2n+1}{n+1}}{2^{2n+1}\,(n+x+1)}=\frac{2^{2x}}{x\,\binom{2x}x}-\frac1x.$$

QUESTION. Let $x>0$ be a real number or an indeterminate. Is this true? $$\sum_{n=0}^{\infty}\frac{\binom{2n+1}{n+1}}{2^{2n+1}\,(n+x+1)}=\frac{2^{2x}}{x\,\binom{2x}x}-\frac1x.$$

POSTSCRIPT. I like to record this presentable form by Alexander Burstein: $$\sum_{n=0}^{\infty}\frac{\binom{2n}n}{2^{2n}(n+x)}=\frac{2^{2x}}{x\binom{2x}x}.$$

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Martin Sleziak
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T. Amdeberhan
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Proving a binomial sum identity

QUESTION. Let $x>0$ be a real number or an indeterminate. Is this true? $$\sum_{n=0}^{\infty}\frac{\binom{2n+1}{n+1}}{2^{2n+1}\,(n+x+1)}=\frac{2^{2x}}{x\,\binom{2x}x}-\frac1x.$$