I am hoping to find out where the formula $$\frac{\pi}{2}=\sum_{k=0}^{\infty}\frac{k!}{\left(2k+1\right)!!}$$ comes from. I can't see how one could begin to prove it.
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Let us start with the geometric series: $$\frac1{1-r}=\sum_{k=0}^\infty r^k$$ If we let $r=-x^2$ and integrate both sides from zero to one, we get the famous Leibniz formula for $\pi$. $$\frac\pi4=\arctan(1)=\int_0^1\frac1{1+x^2}\ dx=\int_0^1\sum_{k=0}^\infty(-x^2)^k\ dx=\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}$$ Applying an Euler Transform, we arrive at $$\begin{align}\frac\pi4&=\sum_{n=0}^\infty\frac1{2^{1+n}}\sum_{k=0}^n\binom nk\frac{(-1)^k}{2k+1}\\\\&=\sum_{n=0}^\infty\text{simplifying the inner sum}\\\\&=\frac12\sum_{n=0}^\infty\frac{k!}{(2k+1)!!}\end{align}$$ The simplifying step comes by noting that $$\sum_{k=0}^0\binom0k\frac{(-1)^k}{2k+1}=\frac11=2^0\frac{0!}{1!!}\color{green}\checkmark$$ $$\sum_{k=0}^1\binom1k\frac{(-1)^k}{2k+1}=\left(\frac11-\frac13\right)=2^1\frac{1!}{3!!}\color{green}\checkmark$$ $$\sum_{k=0}^2\binom2k\frac{(-1)^k}{2k+1}=\left(\frac11-\frac13\right)-\left(\frac13-\frac15\right)=2^2\frac{2!}{5!!}\color{green}\checkmark$$ $$\sum_{k=0}^3\binom3k\frac{(-1)^k}{2k+1}=\left[\left(\frac11-\frac13\right)-\left(\frac13-\frac15\right)\right]-\left[\left(\frac13-\frac15\right)-\left(\frac15-\frac17\right)\right]=2^3\frac{3!}{7!!}\color{green}\checkmark$$ You can prove by induction (and some observation) that the denominators are clearly odd double factorials, and with some work, you can derive the numerators. |
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