SUM INVOLVING THE CENTRAL BINOMIAL COEFFICIENT

Here is another interesting (finite) sum involving the central Binomial Coefficient from this Twitter post:

$\sum _{k=0}^n { n \choose k} \frac{(-1)^k}{2k+1} = \frac{(2n)!!}{(2n+1)!!}$

Proof:

$\begin{aligned} \sum _{k=0}^n { n \choose k} \frac{(-1)^k}{2k+1} &=\sum _{k=0}^n { n \choose k} (-1)^k \int_0^1x^{2k}\,dx\\ &= \int_0^1 \left(\sum _{k=0}^n { n \choose k} (-x^2)^k \right) \,dx\\ &= \int_0^1 \left(1-x^2 \right)^n \,dx\\ &=\frac12 \int_0^1 \left(1-x \right)^n x^{-1/2} \,dx \qquad \left(x^2 \to x \right)\\ &=\frac12\frac{\Gamma\left(n+1\right)\Gamma\left(\frac12\right)}{\Gamma\left(n+1+\frac12\right)}\\ &=\frac12\frac{n\Gamma\left(n\right)\sqrt{\pi}}{\left(n+\frac12 \right)\Gamma\left(n+\frac12\right)}\\ &=\frac12\frac{\Gamma\left(n+1\right)\sqrt{\pi}}{\left(n+\frac12 \right)\frac{(2n-1)!!\sqrt{\pi}}{2^n}}\\ &=\frac{2^n n!}{\left(2n+1 \right)(2n-1)!!}\\ &=\frac{2 n!!}{(2n+1)!!} \qquad \blacksquare\\ \end{aligned}$

$\Gamma\left( n+\frac{1}{2}\right)&=\frac{(2n-1)!!}{2^n}\sqrt{\pi}$

Proof:

$\begin{aligned} \Gamma\left( n+\frac{1}{2}\right)&=\left( n+\frac{1}{2}-1\right)\Gamma\left( n+\frac{1}{2}-1\right)\\ &=\left( n+\frac{1}{2}-1\right)\left( n+\frac{1}{2}-2\right)\Gamma\left( n+\frac{1}{2}-2\right)\\ &= \cdots\\ &=\left( n+\frac{1}{2}-1\right)\left( n+\frac{1}{2}-2\right)\,\cdots \,\left( n+\frac{1}{2}-n\right)\Gamma\left( n+\frac{1}{2}-n\right)\\ &=\left( n-\frac{1}{2}\right)\left( n-\frac{3}{2}\right)\,\cdots \,\frac{1}{2}\,\Gamma\left( \frac{1}{2}\right)\\ &=\left( \frac{2n-1}{2}\right)\left( \frac{2n-3}{2}\right)\,\cdots \,\frac{1}{2}\,\sqrt{\pi}\\ &=\frac{(2n-1)!!}{2^n}\sqrt{\pi} \qquad \blacksquare\\ \end{aligned}$

2n!!=2^n n!

Proof:

$\begin{aligned} 2n!!&=(2n)(2n-2)(2n-4)\cdots4 \cdot 2\\ &=2\times(n)\cdot2\times(n-1)\cdot2\times(n-2)\cdots 2\times 2\cdot 2 \times 1\\ &=2^n n(n-1)(n-2) \cdots 3 \cdot 2 \cdot 1\\ &=2^n n! \qquad \blacksquare \end{aligned}$

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