# Proving that $\sum\limits_{n=0}^{\infty }\frac{1}{(2n)!!}=\sqrt{e}$

Proving that $$\sum_{n=0}^{\infty }\frac{1}{(2n)!!}=\sqrt{e}$$ Firstly, I tried to check the value with the exponential function at $$x=.5$$ but I found its terms not equal to the series terms.

• Could you also prove that $\quad\displaystyle\sum_{n=0}^\infty\frac1{(2n+1)!!}=\sqrt e~\int_0^1e^{-x^2/2}~dx\quad?~$ :-) – Lucian Jan 16 '15 at 19:18

Note that $$(2n)!! = 2\cdot4\cdot 6 \cdots 2n = 2^n n!$$ so your series is just $$\sum_n \frac{(1/2)^n}{n!} = e^{\frac{1}{2}}$$
$$e^x = \sum_{n=0}^\infty\dfrac{x^n}{n!}$$
Plug $x = \dfrac{1}{2}$, then $\dfrac{x^n}{n!} = \dfrac{1}{2^n n!} = \dfrac{1}{2 \cdot 4 \cdot 6 \cdots 2n} = \dfrac{1}{(2n)!!}$