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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.

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    $\begingroup$ Could you also prove that $\quad\displaystyle\sum_{n=0}^\infty\frac1{(2n+1)!!}=\sqrt e~\int_0^1e^{-x^2/2}~dx\quad?~$ :-) $\endgroup$ – Lucian Jan 16 '15 at 19:18
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Note that $$(2n)!! = 2\cdot4\cdot 6 \cdots 2n = 2^n n!$$ so your series is just $$\sum_n \frac{(1/2)^n}{n!} = e^{\frac{1}{2}}$$

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$$e^x = \sum_{n=0}^\infty\dfrac{x^n}{n!}$$

Plug $x = \dfrac{1}{2}$, then $\dfrac{x^n}{n!} = \dfrac{1}{2^n n!} = \dfrac{1}{2 \cdot 4 \cdot 6 \cdots 2n} = \dfrac{1}{(2n)!!}$

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