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I know that :

$$\sum_{n\in\mathbb{N}^*} \binom{2n}{n}\frac{1}{n4^n}= \sum_{n\in\mathbb{N}^*} \frac{(2n)!}{(n!)^2n4^n}=\ln(4)$$

But I have no idea on how to prove it ! Is it a well-known series ?

I need it for a second calculation of a Dirichlet integral :

$$ \int_0^{\frac{\pi}{2}} \ln(\sin(x)) \mathrm{d}x$$

Where I'm using series development and Wallis integral. (I don't need help on this, this is just for context, I'm curious)

Thomas Andrews
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LexLarn
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    If that sum includes an $n=0$ term it diverges, but if by $\Bbb N^\ast$ you meant $n\ge1$ you can write the series as $\int_0^{1/4}\frac{dx}{x}\left((1-4x)^{-1/2}-1\right)$. – J.G. Dec 08 '23 at 14:40
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    This looks vaguely likely the Taylor series for the logarithm. Perhaps it's related via some series acceleration formula? Alternatively, this should follow from the fact $$\sum_{n=0}^\infty \binom{2n}{n} x^{n}= \frac{1}{\sqrt{1-4x}}.$$ – Brevan Ellefsen Dec 08 '23 at 14:49

1 Answers1

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They gave all the hints in the comments.

For $|x|<\frac14$, we have the Maclaurin series $$(1-4x)^{-\frac12}=\sum_{n=0}^{\infty}{-\frac12\choose n}(-4x)^n1^{-\frac12-n}\\ =\sum_{n=0}^{\infty}\frac{(-\frac12)(-\frac32)...(-\frac{2n-1}2)}{n!}(-1)^n4^nx^n\\ =\sum_{n=0}^{\infty}\frac{1.3.5...(2n-1)}{2^n n!}4^nx^n\\ =\sum_{n=0}^{\infty}\frac{(2n)!}{2^n2^n\,n!n!}4^nx^n\\ =\sum_{n=0}^{\infty}{2n\choose n}x^n\tag1 $$ From $(1)$ we can obtain $$\sum_{n=1}^\infty{2n\choose n}x^{n-1}= \frac{(1-4x)^{-\frac12}-1}x.\tag2$$ Integrating $(2)$ from $0$ to $\frac14$ we have $$\sum_{n=1}^\infty{2n\choose n}\frac1{n4^n}=\int_0^{1/4}\left(\frac1{\sqrt{1-4x}}-1\right)\frac{dx}x\\\stackrel{x=\frac{\sin^2\theta}4}{=}2\int_0^{\frac\pi 2}\tan(\tfrac\theta 2)d\theta=2\ln2.$$

Bob Dobbs
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  • I absolutely don't understand the link between $I=\int_0^{1/4}\left(\frac1{\sqrt{1-4x}}-1\right)\frac{dx}x$ and the initial series. But I agree with your equalities. – LexLarn Dec 08 '23 at 18:48
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    THANK YOU for that great edit, I understood everything – LexLarn Dec 08 '23 at 19:45
  • I didnt understand the connection of this with the second calculation. Can you give more details in OP?@LexLarn – Bob Dobbs Dec 08 '23 at 20:15