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I have been looking to evaluate $$\mathcal{A} = \sum_{k=0}^\infty \frac{1}{(2k+1)^3}.$$ We can represent our sum in terms of the Hurwitz zeta function; namely, $$\mathcal{A} = \zeta\left(\frac{1}{2}, 3\right) = \frac{1}{8}\sum_{k=0}^\infty \frac{1}{\left(k+\frac{1}{2}\right)^2}.$$ And from here , we know that
$$\frac{\psi^{\left(-1/2\right)}(3)}{\sqrt{\pi i}} = \zeta\left(\frac{1}{2}, 3\right)$$ which I have no idea how to compute. I am sure there is a less cumbersome way to evaluate this sum. The answer to the sum is $\frac{7}{8}\zeta\left(3\right)$ which seems like it would be a standard computation. Any help would be greatly appreciated.

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$$\zeta(3)=\sum_{n=1}^\infty\frac1{(2n)^3}+\sum_{n=0}^\infty\frac1{(2n+1)^3}=\frac18\sum_{n=1}^\infty\frac1{n^3}+\sum_{n=0}^\infty\frac1{(2n+1)^3}$$

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  • $\begingroup$ Ah, much simpler than my approach. Thank you! $\endgroup$ – user753116 May 3 at 16:58
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Note that for $s>1$ $$\sum_{k\ge0}\frac1{(2k+1)^s}=\sum_{k\ge1}\frac1{k^s}-\sum_{k\ge1}\frac1{(2k)^s}=\sum_{k\ge1}\frac1{k^s}-\frac1{2^s}\sum_{k\ge1}\frac1{k^s}=\left(1-\frac1{2^s}\right)\zeta(3)$$ Set $s=3$ to obtain your result.

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$$ \sum_{k=1}^{\infty}\dfrac{1}{k^3} - \sum_{k =1}^{\infty}\dfrac{1}{(2k)^3}= \left(1 - \dfrac{1}{8}\right)\sum_{k =1}^{\infty}\dfrac{1}{k^3}= \dfrac{7}{8}\zeta(3)$$

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