# Binomial sum of $n$ terms in closed form

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 '13 at 23:30
@Adi: The $2k$ looks like a typo, it should be $2^k$. (Is the calculation correct?) –  TMM May 7 '13 at 22:59
The numerators increase much faster than the denominators, so this won't converge. –  Fred Kline May 12 '13 at 4:11

It is doubtful that a closed form exists for this sum. However, the terms grow very quickly. For $n$ large the sum is well approximated by its last term, $$\sum_{k=1}^n \frac{1}{k}{2^k\choose k-1} \sim \frac{1}{n}{2^n\choose n-1}.$$ For $n=13$ the error due to this approximation is one part in a million. For $n=20$ the error is one part in ten billion.