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A beautiful Gamma series

Let \Gamma denote the Gamma function. Prove that

    \[\sum_{n=0}^{\infty} \frac{\Gamma \left ( n + \frac{3}{2} \right )}{(2n+1)(2n+3) (n+1)!} = -\frac{\pi \sqrt{\pi}}{4} + \sqrt{\pi}\]

Solution

The \arcsin Taylor series is

    \[\arcsin x = \sum_{n=0}^{\infty}{\frac{\left( 2n \right) !}{4^n\left( n! \right) ^2\left( 2n+1 \right)}x^{2n+1}}\]

Hence,

\begin{aligned} \sum\limits_{n=0}^{\infty }{\frac{\Gamma \left( n+\frac{3}{2} \right)}{\left( 2n+1 \right)\left( 2n+3 \right)\left( n+1 \right)!}}&=\sum\limits_{n=1}^{\infty }{\frac{\Gamma \left( n+\frac{1}{2} \right)}{\left( 2n-1 \right)\left( 2n+1 \right)n!}}\\ &=\Gamma \left( \frac{1}{2} \right)+\frac{\sqrt{\pi }}{2}\sum\limits_{n=0}^{\infty }{\frac{\left( 2n \right)!}{{{4}^{n}}{{\left( n! \right)}^{2}}}\cdot \frac{2n+1-\left( 2n-1 \right)}{\left( 2n-1 \right)\left( 2n+1 \right)}\\ &=\sqrt{\pi }+\frac{\sqrt{\pi }}{2}\sum\limits_{n=0}^{+\infty }{\frac{\left( 2n \right)!}{{{4}^{n}}{{\left( n! \right)}^{2}}}\left( \frac{1}{2n-1}-\frac{1}{2n+1} \right)}\\ &=\sqrt{\pi }-\frac{\sqrt{\pi }}{2}\arcsin 1 \\ &=\sqrt{\pi }\left( 1-\frac{\pi }{4} \right)} \end{aligned}

and the result follows.

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