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I stumbled upon the infinite sum $$ 1 + \sum_{k=0}^\infty \frac{(2\pi)^k}{(2k + 1)!!} = 2 + \frac{2\pi}{3} + \frac{(2\pi)^2}{3\cdot 5} + \frac{(2\pi)^3}{3\cdot 5 \cdot 7} + \dots $$ (This is actually the sum of the volume of the unit balls in all odd dimensions.)

From computation, it seems that it converges to something around 12.43. I'm wondering if there's a closed form. Any hints?

(The sum of the volumes of the unit balls in all even dimensions is known to be $\exp(\pi)\approx 23.14$. The sum is the same as above, just with even factors in the denominator.)

Nico Schlömer
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    Take a look at the answer for https://math.stackexchange.com/questions/1482211/double-factorial-series – Stefan Lafon Oct 16 '21 at 17:58
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    According to Mathematica, $$\sum_{k=0}^\infty\frac{(2x)^k}{(2k+1)!_2}=\frac{\sqrt{\pi}}{2}\frac{e^x\operatorname{erf}(\sqrt{x})}{\sqrt{x}}$$ But I am unable to produce a proof. – K.defaoite Oct 16 '21 at 18:09
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    One of the proofs is here -https://math.stackexchange.com/questions/4275413/standard-way-to-evaluate-sum-k-0-infty-fracxkk1-2/4275491#4275491 – Svyatoslav Oct 16 '21 at 19:18

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Thanks for the comments to @K.defaoite and @Svyatoslav.

From here, we have $$ \DeclareMathOperator{\erf}{erf} \sum_{k=0}^\infty \frac{x^k}{(k + 1/2)!} = \sum_{k=0}^\infty \frac{(2x)^k (2k)!!}{(2k + 1)!} = \frac{\sqrt{\pi}}{2} \frac{\exp(x)}{\sqrt{x}} \erf{\sqrt{x}}, $$ so $$ \sum_{k=0}^\infty \frac{(2\pi)^k}{(2k + 1)!!} = \frac{1}{2} \exp(\pi) \erf(\sqrt{\pi}) $$ which indeed is $12.429316728301792$...

This means that the sum of the volumes of the unit balls in all dimensions is $$ \begin{align} \sum_{n=0}^\infty |S_n| &= \sum_{n=0}^\infty |S_{2n}| + \sum_{n=0}^\infty |S_{2n + 1}|\\ &= \exp(\pi) + \left(1 + \frac{1}{2} \exp(\pi) \erf(\sqrt{\pi})\right)\\ &= 1 + \exp(\pi) \left(1 + \frac{1}{2} \erf(\sqrt{\pi})\right)\\ &\approx 35.57 \end{align} $$

Nico Schlömer
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