Closed form of $$\sum_{n=0}^{\infty }\frac{1}{2^n(4n-1)}$$
the Wolfram gave me the answer below, but I ask if there is a clear value of this sum
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E.H.E
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$\frac{\coth ^{-1}\left(\sqrt[4]{2}\right)-\cot ^{-1}\left(\sqrt[4]{2}\right)}{2 \sqrt[4]{2}}-1$ – Pierpaolo Vivo Jan 18 '16 at 21:38
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Or, if it is more useful $-1+\int_0^{\infty } \frac{\exp (s)}{2 e^{4 s}-1} , ds$ – Pierpaolo Vivo Jan 18 '16 at 21:41
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@PierpaoloVivo, thanks, but how did you get it? – E.H.E Jan 18 '16 at 21:44
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1separate the $n=0$ contribution. Then, in the summation from $1$ to $\infty$ use $\frac{1}{4n-1}=\int_0^\infty ds\ e^{-s (4n-1)}$ and exchange integral and summation. – Pierpaolo Vivo Jan 18 '16 at 21:45
2 Answers
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We have
$$\begin{align} \sum_{n=0}^\infty \frac{1}{2^n(4n-1)}&=-1+\sum_{n=1}^\infty \frac{1}{2^n}\int_0^1 x^{4n-2}\,dx\\\\ &=-1+\sum_{n=1}^\infty \int_0^1 \frac{(x^4/2)^n}{x^2}\,dx\\\\ &=-1+\int_0^1\frac{x^2}{2-x^4}\,dx\\\\ &=-1+\frac{\text{atanh}\left(2^{-1/4}\right)-\text{atan}\left(2^{-1/4}\right)}{2^{5/4}}\\\\ &\approx -0.779247 \end{align}$$
Mark Viola
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@E.H.E Yes it is. Edited! And much appreciative of your catching the typo. +1 – Mark Viola Jan 18 '16 at 21:50
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Consider
$$f(x) = \sum_{n=0}^{\infty} \frac{x^{4 n}}{4 n+3} $$
$$\frac{d}{dx} \left [x^3 f(x) \right ] = \sum_{n=0}^{\infty} x^{4 n+2} = \frac{x^2}{1-x^4}$$
Integrate both sides...
$$x^3 f(x) = \frac14 \log{\left (\frac{1+x}{1-x} \right )} - \frac12 \arctan{x} + C $$
where $C=0$ by considering $x=0$. The series is equal to
$$f(2^{-1/4}) = \sum_{n=0}^{\infty} \frac1{2^n} \frac1{4 n+3} = 2^{3/4} \left [\frac14 \log{\left (\frac{1+2^{-1/4}}{1-2^{-1/4}} \right )} - \frac12 \arctan{2^{-1/4}} \right ]$$
The relation to your series is
$$-1 + \frac12 f(2^{-1/4}) \approx -0.779247$$
Ron Gordon
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