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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

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  • $\begingroup$ Does x!! denote the factorial of x!? $\endgroup$ – Guy Mar 6 '14 at 10:33
  • $\begingroup$ Doesn't converge, according to wolframalpha.com/input/… $\endgroup$ – Guy Mar 6 '14 at 10:35
  • $\begingroup$ @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) $\endgroup$ – gammatester Mar 6 '14 at 10:39
  • $\begingroup$ @gammatester my bad. $\endgroup$ – Guy Mar 6 '14 at 10:39
  • $\begingroup$ 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. $\endgroup$ – Guy Mar 6 '14 at 10:43
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Hint: by Binomial theorem or Taylor expansion:

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

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  • $\begingroup$ Ah the infamous binomial theorem. $\endgroup$ – Guy Mar 6 '14 at 10:48

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