Proving that $ \sum_{k=0}^\infty\frac1{2k+1}{2k \choose k}^{-1}=\frac {2\pi}{3\sqrt{3}} $

Question

Mathematica gives away the interesting sum: $$\sum_{k=0}^\infty\frac1{2k+1}{2k \choose k}^{-1}=\frac{2\pi}{3\sqrt{3}}$$ The question is: How to prove it by hand?

Remark. This question is self-answered (whence the OP provided an effort). See the answer below.

$$\sum _{k=0}^{\infty } \frac{z^k}{(2 k+1) \binom{2 k}{k}}=\frac{4 \sin ^{-1}\left(\frac{\sqrt{z}}{2}\right)}{\sqrt{(4-z) z}}$$ — Infiniticism, Aug 22 '20 at 06:40
A simple generalization is a formula due to Ramanujan: $$\sum _{k=0}^{\infty } \frac{1}{(2 k+1)^2 \binom{2 k}{k}}=\frac{1}{3} \left(8 C-\pi \log \left(\sqrt{3}+2\right)\right)$$ — Infiniticism, Aug 22 '20 at 06:41
Someone should prove this (or explain a bit) Ramanujan thing mentioned by @User628759 — Paramanand Singh, Aug 23 '20 at 03:02
Check this solution by a friend https://math.stackexchange.com/a/3800245/432085 — Ali Olaikhan, Aug 24 '20 at 20:00
@Ali Shather Very interesting, in fact, I was already bothering about it in a long way! — Z Ahmed, Aug 25 '20 at 04:33
@Paramanand Singh We have
$$\frac{\arcsin(x)}{\sqrt{1-x^2}}=\sum_{n=0}^\infty \frac{(2x)^{2n+1}}{(2n+1){2n\choose n}}$$

Divide both sides by $2x$ then $\int_0^{1/2}$ we get

$$\sum_{n=0}^\infty \frac{1}{(2n+1)^2{2n\choose n}}=2\int_0^{1/2}\frac{\arcsin(x)}{x\sqrt{1-x^2}}dx=2\int_0^{\pi/6}\frac{\theta}{\sin\theta}d\theta$$

which is well-known — Ali Olaikhan, Aug 25 '20 at 06:18
@AliShather: that's cool! So Ramanujan stuff is not necessarily difficult. — Paramanand Singh, Aug 25 '20 at 07:43

score 5 · Accepted Answer · edited Aug 29 '20 at 13:37

5

I think this might help.

$$\frac{1}{n \choose k}=(n+1)\int_0^1 x^k(1-x)^{n-k}dx.$$ This is easily obtained by integration by parts.

Putting $n=2k$ and $x=\sin^2t$, we get $$\sum_{k\geq0}\int_0^{\frac{\pi}{2}}2\sin^{2k+1}t\cos^{2k+1}t \ dt.$$

Bringing the sum inside the integral and evaluating the G.P. give $$\int_0^{\frac{\pi}{2}}\frac{\sin( 2t)}{1-\frac{\sin^2(2t)}{4}}dt.$$

Putting $\cos(2t)=u$ We get, $$2\int_{-1}^{1}\frac{du}{3+u^2},$$ which is indeed equal to $\dfrac{2\pi}{3\sqrt3}$.

edited Aug 29 '20 at 13:37

Batominovski

50,341

answered Aug 22 '20 at 08:12

LM2357

4,203

Will use that trick [+1] – openspace Aug 29 '20 at 13:33

score 5 · Answer 2 · edited Aug 29 '20 at 13:36

Note the integral representation of the reciprocal of the binomial co-efficient: $${n \choose j}^{-1}=(n+1)\int_{0}^{1} x^j (1-x)^{n-j}~ dx.$$ Then $$S=\sum_{k=0}^{\infty} \frac{1}{2k+1}{2k \choose k}^{-1}= \int_{0}^{1} \sum_{k=0}^{\infty} [x(1-x)]^{k}= \int_{0}^{1} \frac{1}{x^2-x+1} dx $$ $$ \implies S = \frac{2}{\sqrt{3}}\tan^{-1} \frac{2x-1}{\sqrt{3}}\Bigg|_{0}^{1}=\frac{2\pi}{3\sqrt{3}}.$$

Proving that $ \sum_{k=0}^\infty\frac1{2k+1}{2k \choose k}^{-1}=\frac {2\pi}{3\sqrt{3}} $

2 Answers2

Linked