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My problem is to evaluate $\sum_{n=1}^{\infty}(\frac{2}{9})^n\frac{(2n)!}{(n+1)!n!}$. After manually calculating, I am observing that the value seems to approach $\frac{3}{2}$. Peculiar, but I have no idea on how to go about proving that.

Some observations:

  • $\frac{(2n)!}{(n+1)!n!}$ is the $n$th Catalan number.
  • $\sum_{n=1}^{\infty}(\frac{3}{9})^n\frac{(2n)!}{(n+1)!n!}$ diverges to infinity.

I would be astonished if the value of the summation can be calculated. Thank you!

SunnyMoon
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1 Answers1

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You can use the generating function for the central binomial coefficients:

$$\sum_{n=0}^\infty\binom{2n}n x^n = \frac 1{\sqrt{1-4x}} \text{ for } |x| < \frac 14$$

Now, note that $\frac 1{n+1}\binom {2n}n = \frac{(2n)!}{(n+1)!n!}$. So you can rewrite your sum as follows and integrate: $$ \sum_{n=1}^\infty x^n \frac{(2n)!}{(n+1)!n!} = \frac 1x \sum_{n=1}^\infty \frac{x^{n+1}}{n+1}\binom{2n}n= \frac 1x \int_0^x\frac 1{\sqrt{1-4t}}dt$$ $$= \boxed{\frac 1{2x}\left(1-\sqrt{1-4x}\right)}$$

For your $x= \frac 29$ you get the "desired" value of $\frac 32$.

trancelocation
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