A closed form of $\sum_{n=0}^\infty\frac{n!!}{n!}$

Asked Feb 01 '17 at 13:03

Active Feb 01 '17 at 13:03

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It's fairly easy to see, using the ratio test, that the series

$$\sum_{n=0}^\infty\frac{n!!}{n!}$$

converges, where $n!!$ is the double factorial.

However, I'd be interested if there is a way of calculating its value analytically. I've tried around a little with WolframAlpha and I know that

$$\sum_{n=0}^\infty\frac{n!!}{n!}\approx 4.05941$$

Wolfram also suggests a couple of possible closed forms, which look very different. Since this sum is related to $e$ somehow, the following two are what I'd find most likely:

$$\sum_{n=0}^\infty\frac{n!!}{n!}=-8-e+2e^2$$ $$\sum_{n=0}^\infty\frac{n!!}{n!}=\Gamma(e+1)+e-2-\frac{5}{2e}$$

However, I have no clue if either of the two is actually correct, and if it were, how one would go about proving this result to be true - none of the approaches to series that I know seem to be getting me anywhere.

Is there a way to find a closed form for this series, and if so, how would one go about deriving it?

asked Feb 01 '17 at 13:03

Tom

3,289

1

Hint: $n!=n!!(n-1)!!$. – Paul Feb 01 '17 at 13:05
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This looks related: http://math.stackexchange.com/a/169577/42969. – Martin R Feb 01 '17 at 13:06
@MartinR That just answered the question. (I think). – S.C.B. Feb 01 '17 at 13:07
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Or one of these http://math.stackexchange.com/questions/1482211/double-factorial-series, http://math.stackexchange.com/questions/1821138/proof-of-a-formula-containing-double-factorial ? – Martin R Feb 01 '17 at 13:10
@MartinR I think you should answer the question with those links. If I did, that would be cheating. – S.C.B. Feb 01 '17 at 13:12
Thanks a lot for the links, very interesting. @MartinR, if you put these links in an answer, I'll accept it. – Tom Feb 01 '17 at 13:12

A closed form of $\sum_{n=0}^\infty\frac{n!!}{n!}$

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