Combinatorial Sum

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#1 h Apr 24, 2025, 1:02 AM

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Evaluate $\sum_{n=0}^{\infty} \frac{2^n + 1}{(2n + 1) \binom{2n}{n}}$

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#2 h May 3, 2025, 12:56 PM • 1 Y

Y by P162008

$\sum_{n=0}^{\infty} \frac{2^n + 1}{(2n + 1) \binom{2n}{n}} = \sum_{n=0}^{\infty} \frac{2^n + 1}{\binom{2n}{n}}\cdot \int_{0}^{1} x^{2n} dx = \int_{0}^{1} \sum_{n=0}^{\infty} \frac{(\sqrt{2}x)^{2n} + x^{2n}}{\binom{2n}{n}} dx$ ;

or $\int_{0}^{1} \frac{4x \cdot \arcsin(x/2) +4\sqrt{4-x^2}}{(4-x^2)^{3/2}} + \frac{2x \cdot \arcsin(x/\sqrt{2})+2\sqrt{2-x^2}}{(2-x^2)^{3/2}} dx$ ;

or $\frac{4\arcsin(x/2)}{\sqrt{4-x^2}} + \frac{2\arcsin(x/\sqrt{2})}{\sqrt{2-x^2}}|_{0}^{1} = \textcolor{red}{\frac{(9+4\sqrt{3})\pi}{18}}$ .

This post has been edited 5 times. Last edited by Mathzeus1024, May 3, 2025, 1:10 PM

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#3 h May 4, 2025, 1:19 AM • 1 Y

Y by Mathzeus1024

P162008 wrote:

Evaluate $\sum_{n=0}^{\infty} \frac{2^n + 1}{(2n + 1) \binom{2n}{n}}$

$\frac1{(2n+1)\binom{2n}{n}}=\int_{0}^{1}\left(t(1-t)\right)^ndt$

So the given sum is $\int_{0}^{1}\sum_{n \geq 0} \left(\left(2t(1-t)\right)^n + \left(t(1-t)\right)^n \right)dt=\int_{0}^{1}\frac1{1-2t+2t^2}dt +\int_{0}^{1}\frac1{1-t+t^2}dt=\frac{\pi}{2}+\frac{2\pi}{3\sqrt{3}}$

( $2t(1-t), t(1-t)<1$ for $0 \leq t \leq 1$ )

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