How find the $\sum_{n=0}^{\infty}\frac{(2n-1)!!}{(2n)!!}\cdot\left(\frac{1}{2^n}\right)$

Find the sum $$\sum_{n=0}^{\infty}\dfrac{(2n-1)!!}{(2n)!!}\cdot\left(\dfrac{1}{2^n}\right)$$

we know $$(2n-1)<2n$$ so $$\dfrac{(2n-1)!!}{(2n)!!}\cdot\dfrac{1}{2^n}<\dfrac{1}{2^n}$$ so this sum is converge I think use $\arcsin{x}$,But I can't,Thank you

• Does x!! denote the factorial of x!? – Guy Mar 6 '14 at 10:33
• Doesn't converge, according to wolframalpha.com/input/… – Guy Mar 6 '14 at 10:35
• @Sabyasachi: $x!!$ is the double factorial, and according to WA is converges to $\sqrt{2}$. You entered the wrong formula, try sum((2n-1)!!/((2n)!!)/2^n) – gammatester Mar 6 '14 at 10:39
• @gammatester my bad. – Guy Mar 6 '14 at 10:39
• If $$\dfrac{(2n-1)!!}{(2n)!!}\cdot\dfrac{1}{2^n}<\dfrac{1}{2^n}$$ holds, then the summations must be less than 1, which it clearly isn't according to wolfram. I think the error is in the fact that $(-1)!$ isn't defined. – Guy Mar 6 '14 at 10:43

$\sum \frac{(2n-1)!!}{(2n)!!} x^n=\frac{1}{\sqrt{1-x}}, \quad |x|<1$.