So I was was messing around with infinite series and I came across one that is deceptively similar to the familiar $\sum_{n=1}^\infty \frac{1}{2^n} = 1 $ Simply add $1$ to each denominator of each term of the series. This is $\sum_{n=1}^\infty \frac{1}{2^n+1} \approx 0.76449...$ I have used all my weaponry on trying to crack it and the best I've come up with is the equivalent series $\sum_{n=1}^\infty \frac{(-1)^n}{2^n-1} $ Could somebody help me retrieve some sort of compact value for this expression ( e.g. $\sum_{n=1}^\infty \frac{1}{n^2} $ can be rewritten as $\frac{\pi^2}{6}$) or is this pretty much impossible? Thanks!
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Have a look at http://math.stackexchange.com/questions/1978310/find-sum-k-1-infty-frac12k1-1 – Jack D'Aurizio Jun 08 '18 at 20:02
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If you know a closed form for $,\displaystyle f(x):=\prod\limits_{n=1}^\infty \left(1+\frac{x}{2^n}\right),$ (I assume one cannot find one) you can use $,\displaystyle \left(\frac{d}{dx}\ln f(x)\right)|_{x=1},$ to get a "compact value" . – user90369 Jun 10 '18 at 10:59
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There is no known closed form for this sum.
Robert Israel
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None known by me, anyway... But I'm pretty sure this has come up before. – Robert Israel Jun 08 '18 at 18:51
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@ThomKiwi Using wolfram alpha, the result is $-1+\frac{\psi_{\frac{1}{2}}^{(0)}\left(1-\frac{i \pi}{\log(2)}\right)}{\log(2)}$. So i doubt that it has a nice form, like the $\sum\limits_{n \in \mathbb{N}}\frac{1}{n^2}$. – Botond Jun 08 '18 at 19:04
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You could use the fact that $$\frac{1}{2^n+1}+\frac{1}{2^n(2^n+1)} = \frac{1}{2^n},$$ and then attempt to find a closed form for the following sum, which appears to be equally difficult,
$$\sum_{n=1}^\infty \frac{1}{2^n(2^n+1)}$$ and then subtract the closed form of the previous sum from $1.$
Chickenmancer
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