Why does $\sum_{n=0}^\infty \frac{n!}{(2n+1)!!}$ converge to $\frac{\pi}{2}$

Question

Why does:

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

converge to $\frac{\pi}{2}$?

I'm honestly not sure where to start.

Answer 1

$$\sum_{n\geq 0}\frac{n!}{(2n+1)!!}=\sum_{n\geq 0}\frac{2^n n!^2}{(2n+1)\cdot (2n)!}=\sum_{n\geq 0}\frac{2^n}{(2n+1)\binom{2n}{n}}=\sum_{n\geq 0}\frac{2^n \Gamma(n+1)^2}{\Gamma(2n+2)}$$ can be written, through Euler's Beta function, also as $$ \sum_{n\geq 0}2^n\int_{0}^{1}x^n(1-x)^n\,dx = \int_{0}^{1}\frac{dx}{1-2x(1-x)}\stackrel{\text{symmetry}}{=}2\int_{0}^{1/2}\frac{2\,dx}{1+(2x-1)^2} $$ or, through the substitution $x=\frac{1-z}{2}$, as $$ \int_{0}^{1}\frac{2\,dz}{1+z^2}=2\arctan(1)=\color{red}{\frac{\pi}{2}}.$$

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As an alternative, you may notice your series (related to the Taylor series of the arcsine function) is half the Euler transform of Gregory series. See pages 20-21 of my notes, for instance.

Answer 2

1. Apply Euler transform to Leibniz fomula for $\pi$

Leibniz fomula is $$ 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + ~..~ = \frac{\pi}{4} $$

Euler transform is $$ a_0 - a_1 + a_2 - a_3 + ~..~ = \frac{a_0}{2} - \frac{\Delta a_0}{2^2} + \frac{\Delta^2 a_0}{2^3} - \frac{\Delta^3 a_0}{2^4} + ~.. $$ where $\Delta a_0 = a_1 - a_0$, $\Delta^2 a_0 = (a_2 - a_1) - (a_1-a_0) $,$\Delta^3a_0 = ((a_3-a_2)-(a_2-a_1)) - ((a_2-a_1)-(a_1-a_0))$, ..etc.

Proof: Consider shift operator $T a_0 = a_1$, and difference operator $\Delta a_0 = a_1-a_0$. Then $\Delta = T-1$. Therefore \begin{eqnarray} a_0 - a_1 + a_2 - a_3 + ~..~ &=& (1-T+T^2-T^3+~..)a_0 \\ &=& \frac{1}{1+T} a_0 = \frac{1}{2+\Delta} a_0\\ &=& \left(\frac{1}{2} - \frac{\Delta}{2^2} + \frac{\Delta^2}{2^3} - ~..\right) a_0 \end{eqnarray}

Applying this to Leibniz fomula. \begin{eqnarray} a_0 &=& 1 \\ \Delta a_0 &=& - \frac{2 }{1\cdot 3} \\ \Delta^2 a_0 &=& \frac{2 \cdot 4}{1\cdot 3 \cdot 5} \\ \Delta^3 a_0 &=& - \frac{2 \cdot 4 \cdot 6}{1\cdot 3 \cdot 5 \cdot 7} \\ .. \end{eqnarray}

\begin{eqnarray} 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + ~..~ = \frac{1}{2} \left(1 + \frac{1}{3} + \frac{1 \cdot 2}{3 \cdot 5} + \frac{1 \cdot 2 \cdot 3}{3 \cdot 5\cdot 7} + ~..\right) = \frac{\pi}{4} \end{eqnarray} Thus $$ \sum_{n=0}^\infty \frac{n!}{(2n+1)!!} = \frac{\pi}{2} $$

2. Use beta function related method. See $$ \int_0^1 (1-x^2)^n dx = \frac{2^n n!}{(2n+1)!!} $$ Proof: \begin{eqnarray} (x (1-x^2)^n)' &=& (1-x^2)^n - 2n x^2(1-x^2)^{n-1} \\ &=& (1-x^2)^n + 2n (1-x^2)^{n-1} - 2n x^2(1-x^2)^{n-1} - 2n (1-x^2)^{n-1} \\ &=&(1+2n)(1-x^2)^n - 2n (1-x^2)^{n-1} \end{eqnarray} Therefore \begin{eqnarray} \int_0^1 (1-x^2)^n dx = \frac{2 n}{2n+1}\int_0^1 (1-x^2)^{n-1} dx \end{eqnarray} Then the result readily follows. Then

\begin{eqnarray} \sum_{n=0}^\infty \frac{n!}{(2n+1)!!} &=& \int_0^1 dx \left(1 + \frac{1-x^2}{2} + \frac{(1-x^2)^2}{4} + \frac{(1-x^2)^3}{8} + ~..\right) \\ &=& \int_0^1 \frac{1}{1-\frac{1-x^2}{2}} dx \\ &=& \int_0^1 \frac{2dx}{1+x^2} = \frac{\pi}{2} \end{eqnarray}

Answer 3

What about finding a function $f$ for which there exists an $x$ such that $f(x)=\frac{\pi}{2}$, calculating its Taylor series and using the Euler Transform for series?

Start with $a\tan(1)=\frac{\pi}{4}$ and search for the Taylor series of $a\tan(x)$: $$a\tan(x)=f(x) = \sum _{n=0}^{\infty } \frac{(-1)^n}{(2n+1)}x^n.$$

Looking at the Euler transform gives us $$ \left(\frac{1}{1-x}\right)f\left(\frac{x}{1-x}\right) = \sum _{n=0}^{\infty } \left(\sum _{k=0}^n {n \choose k}\frac{(-1)^k}{(2k+1)}\right)x^n.$$

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

We see that the series $$ \sum_{n=0}^{\infty} \frac{n!}{(2n+1)!!} =\sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!} \left(\frac{1}{2}\right)^n = 2f(1)=\frac{\pi}{2}$$