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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

math110
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  • Does x!! denote the factorial of x!? – Guy Mar 06 '14 at 10:33
  • Doesn't converge, according to http://www.wolframalpha.com/input/?i=sum((2n-1)!!%2F(2n!!)%2F2%5En)+from+0+to+infinity – Guy Mar 06 '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 06 '14 at 10:39
  • @gammatester my bad. – Guy Mar 06 '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 06 '14 at 10:43
  • @Sabyasachi $(-1)!!=1$ by convention. – Ma Ming Mar 06 '14 at 10:45
  • @MaMing yeah it's okay. Actually I made a calculation error, thinking, $$\sum_0^\infty \frac{1}{2^n} = 1$$ while infact it is 2. – Guy Mar 06 '14 at 10:47

1 Answers1

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Hint: by Binomial theorem or Taylor expansion:

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

Ma Ming
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