Timeline for Closed form for $\sum_{n=1}^\infty \frac{4^n}{n^p\binom{2n}{n}}$
Current License: CC BY-SA 4.0
15 events
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Jun 30, 2019 at 15:18 | comment | added | Tito Piezas III | You may like this related post. | |
Jun 3, 2019 at 15:57 | history | edited | Robert Z | CC BY-SA 4.0 |
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Jun 3, 2019 at 15:51 | history | edited | Robert Z | CC BY-SA 4.0 |
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Jun 3, 2019 at 15:16 | history | edited | Robert Z | CC BY-SA 4.0 |
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Jun 3, 2019 at 13:48 | history | edited | Robert Z | CC BY-SA 4.0 |
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Jun 3, 2019 at 12:57 | history | edited | Robert Z | CC BY-SA 4.0 |
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Jun 3, 2019 at 12:24 | comment | added | Robert Z | So we have some kind of closed form. See my edit. | |
Jun 3, 2019 at 12:19 | history | edited | Robert Z | CC BY-SA 4.0 |
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Jun 3, 2019 at 11:51 | comment | added | Zacky | For $p=4$ we easily get $$\sum_{n=1}^\infty \frac{4^n}{n^4\binom{2n}{n}}=-8\int_0^\frac{\pi}{2} x\ln^2(\sin x)dx$$ So I can bet there will be a closed form. | |
Jun 3, 2019 at 11:29 | history | edited | Robert Z | CC BY-SA 4.0 |
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Jun 3, 2019 at 11:23 | history | edited | Robert Z | CC BY-SA 4.0 |
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Jun 3, 2019 at 11:06 | comment | added | Claude Leibovici | Thanks for the link and $\to +1$ | |
Jun 3, 2019 at 11:04 | history | edited | Robert Z | CC BY-SA 4.0 |
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Jun 3, 2019 at 10:56 | history | edited | Robert Z | CC BY-SA 4.0 |
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Jun 3, 2019 at 10:50 | history | answered | Robert Z | CC BY-SA 4.0 |