A while ago I had a dream, the series you see below appeared in front of my eyes
$$\sum_{n=1}^{\infty} \frac{1}{n 2^n (2^n)! }$$
Do you think it is possible to find a closed form of it?
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6Does Goddess Namagiri whisper this series to you in your dream? – Tunk-Fey Sep 06 '14 at 12:42
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@Tunk-Fey No. I also dreamt this series before meeting it in my research $$\sum_{n=0}^{\infty} \arcsin\left(\frac{1}{e^n}\right)$$ – user 1591719 Sep 06 '14 at 12:50
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1What did you consume before going to sleep? – copper.hat Sep 06 '14 at 12:57
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It obviously converges. The terms get small quickly - the 4th term is less than $10^{-15}$ and the 5th less than $10^{-37}$. Isn't an approximation to 3 terms good enough? – almagest Sep 06 '14 at 12:58
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ISC and WA do not yield any "feasible" results. – Ali Caglayan Sep 06 '14 at 13:32
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Are you sure it was a dream ? – Claude Leibovici Sep 06 '14 at 13:34
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ISC reports nothing for 0.2552093667328 so certainly it is not a "well known" closed form. I think "the number is transcendental" is more likely to be proved than a "closed form" for it. – GEdgar Sep 06 '14 at 14:01
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You should post $$\sum_{n=0}^{\infty} \arcsin\left(\frac{1}{e^n}\right)$$ It is a very nice one. May I wish you nice and good dreams (don't come back with another nightmare, please !). Cheers :-) – Claude Leibovici Sep 06 '14 at 14:29
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@ClaudeLeibovici :-))) – user 1591719 Sep 06 '14 at 14:38
1 Answers
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Do you think it is possible to find a closed form of it?
No. I don't. If for no other reason, then merely because I am utterly unaware of any non-trivial
series of the form $\displaystyle\sum_{n=\ldots}^\infty\frac{\ldots}{\ldots(a^n)!\ldots}$ to have a known closed form.
Lucian
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1@Chris'ssis: And one day we might. :-) But, to my knowledge, that day is yet to come. – Lucian Sep 06 '14 at 20:11