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I wonder if there is a simple closed form solution to the following sum: $\sum_{k = 1}^n \frac{(1/2)^k}{k}$? Wolfram Alpha expresses it in terms of the Lerch transcendent, but I wonder if there is a more simple formula? Thanks.

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Is the numerator of your summand supposed to be $(1/2)^k$? – Nick Gill Feb 5 at 11:11
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 $\log(2)$ and this is not a reseach-level question. – Andrej Bauer Feb 5 at 11:12
Yes the numerator is $(1/2)^k$. @Andrej: Note that the sum does not go to infinity, but to a finite integer $n$. – Danne Feb 5 at 11:24
Maple uses Lerch too. – Brendan McKay Feb 5 at 11:40
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 If there were a simple closed expression for $\sum_{k=1}^n\frac{x^k}{k}$, then there were a simple expression for the partial harmonic series coming from $x=1$, and we know that such a thing does not exist. – Peter Mueller Feb 5 at 13:05
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closed as off topic by Andrej Bauer, Chris Godsil, Nik Weaver, Anthony Quas, Andy Putman Feb 6 at 3:29

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