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Can we calculate the given combinatorial sum in closed form?

$$ \frac{\binom{2}{0}}{1}+\frac{\binom{4}{1}}{2}+\frac{\binom{8}{2}}{3}+\frac{\binom{16}{3}}{4}+\cdots+\frac{\binom{2^n}{n-1}}{n}$$

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According to Wolfram Mathematica $S_n=\sum_{k=1}^{n}\frac{1}{k}\binom{2k}{k-1}=2(-2 + (e^{2 + n}\Gamma(2 + n, 1)/\Gamma(3 + n))$ – Adi Dani Jan 28 at 23:30
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@Adi: The $2k$ looks like a typo, it should be $2^k$. (Is the calculation correct?) – TMM May 7 at 22:59
The numerators increase much faster than the denominators, so this won't converge. – Fred Kline May 12 at 4:11

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