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Given a sequence $\{x_n\}, \ n\in\Bbb N$: $$ x_n = \frac{(2n)!!}{(2n+1)!!} $$ Show that $x_n$ converges.

I'm wondering why I'm getting a seemingly wrong result (assuming the problem statement asks to prove convergence): $$ \begin{align} x_n &= \frac{(2n)!!}{(2n+1)!!} \\ &= \frac{2\cdot 4\cdot 6\cdots (2n-2)\cdot(2n)}{3\cdot 5\cdot 7\cdots (2n-1)\cdot(2n+1) } \\ &= \frac{2\cdot 4\cdot 6\cdots (2n-2)\cdot(2n)}{3\cdot 5\cdot 7\cdots (2n-1)\cdot(2n+1)} \cdot \frac{2(n)!}{2(n)!} \\ &= \frac{4^n (n!)^2}{(2n+1)!} \\ &= \frac{4^n (n!)^2}{(2n+1)\cdot (2n)!} \\ &=\frac{4^n}{2n+1}\cdot \frac{(n!)^2}{(2n)!} \end{align} $$

By Binomial coefficients: $$ {2n\choose n} = \frac{(2n)!}{n!(2n-n)!} = \frac{(2n)!}{(n!)^2} $$

Thus: $$ x_n = \frac{4^n}{2n+1}\cdot \frac{1}{{2n \choose n}} $$

Doesn't $\frac{4^n}{2n+1}$ grow faster than $\frac{1}{2n\choose n}$ is declining? Shouldn't $x_n$ diverge in that case?

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  • $\begingroup$ I don't see how $$\frac{2\cdot 4\cdot 6\cdots (2n-1)\cdot(2n)}{3\cdot 5\cdot 7\cdots (2n-1)\cdot(2n+1)} \cdot \frac{(2n)!}{(2n)!} = \\= \frac{4^n (n!)^2}{(2n+1)!} $$ would be true. $\endgroup$ – 5xum Jan 3 at 13:05
  • $\begingroup$ The asymptotic behavior of $\binom {2n} n$ could be determined by Stirling approximation. It may not be slow as you think. $\endgroup$ – xbh Jan 3 at 13:05
  • $\begingroup$ @5xum I made a mistake in the nominator. $\endgroup$ – roman Jan 3 at 13:08
  • $\begingroup$ @5xum, there is a mistake in the notation, it should be 2(n!) instead of (2n)! to make this result true. You can then compute this using Stirling. $\endgroup$ – S. Franssen Jan 3 at 13:12
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You have

$$0\leq x_{n+1} = x_n \cdot \frac{2n+2}{2n+3} \leq x_n$$ thus, your sequence is monotone decreasing and bounded from below and therefore convergent.

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Using the Stirling approximation for factorials we can observe the following: $n! \sim \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n$. This means that, while being a bit sloppy with the $\sim$

$ \begin{align*} x_n &= \frac{(2n)!!}{(2n+1)!!} \\ &= \frac{4^n (n!)^2}{(2n+1) (2n)!}\\ & \sim \frac{4^n 2 \pi n \left( \frac{n}{e} \right)^{2n}}{ (2n+1) 2\sqrt{\pi n} \left(\frac{2n}{e} \right)^{2n}}\\ &= \frac{\sqrt{\pi n}}{2 n + 1} \rightarrow 0 \end{align*} $

So $x_n \rightarrow 0$

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