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Does the series $\displaystyle\sum^\infty_{n=1}\frac{n!}{\sqrt{(2n)!}}$ converge/diverge?

I used the ratio test but I'm not sure:

$\begin{align} \frac{\frac{(n+1)!}{\sqrt{(2n+2)!}}}{\frac{n!}{\sqrt{(2n)!}}} &=\frac{n+1}{\sqrt{(2n+1)(2n+2)}}\\ &=\frac{n+1}{4(n+1)^2\sqrt{2n+1}}\\ &=\frac{1}{4\sqrt{2n+1}} \end{align}$

The limit of that is smaller than $1$ so the series does converge. Is it correct ?

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    $\begingroup$ There is an error in the second line of your calculation. The limit you should get is $1/2$. But yes, the conclusion is still correct. $\endgroup$ – Antonio Vargas Dec 21 '13 at 21:34
  • $\begingroup$ @AntonioVargas I can't take them out of the root like so $\frac{n+1}{4(n+1)^2\sqrt{2n+1}}$ ? That's what I wasn't sure about. $\endgroup$ – Senishoshitsu Dec 21 '13 at 21:38
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This is in response to your question here. It's too long for a comment so I'm posting it as an answer.

No, you can't take them out of the root like that. One way to handle the denominator is by writing it like

$$ \begin{align} \sqrt{(2n+1)\cdot(2n+2)} &= \sqrt{(2n+1)\cdot 2\cdot (n+1)} \\ &= \sqrt{2n+1} \sqrt{2} \sqrt{n+1}, \end{align} $$

so that

$$ \frac{n+1}{\sqrt{(2n+1)(2n+2)}} = \frac{n+1}{\sqrt{2} \sqrt{2n+1} \sqrt{n+1}}. $$

Now

$$ \frac{n+1}{\sqrt{n+1}} = \sqrt{n+1}, $$

so

$$ \frac{n+1}{\sqrt{2} \sqrt{2n+1} \sqrt{n+1}} = \frac{\sqrt{n+1}}{\sqrt{2}\sqrt{2n+1}}. $$

To calculate the limit of this as $n \to \infty$, divide the numerator and the denominator by $\sqrt{n}$ to get

$$ \begin{align} \frac{\frac{1}{\sqrt{n}}\sqrt{n+1}}{\frac{\sqrt{2}}{\sqrt{n}}\sqrt{2n+1}} &= \frac{\sqrt{1+\frac{1}{n}}}{\sqrt{2}\sqrt{2 + \frac{1}{n}}} \\ &\overset{n\to\infty}{\longrightarrow} \frac{\sqrt{1+0}}{\sqrt{2}\sqrt{2 + 0}} \\ &= \frac{1}{2}. \end{align} $$

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    $\begingroup$ Woah community wiki, thanks for clearing that out. $\endgroup$ – Senishoshitsu Dec 21 '13 at 22:02
  • $\begingroup$ You're welcome :) $\endgroup$ – Antonio Vargas Dec 22 '13 at 0:51
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Hint:

$$\left( \frac{n!}{\sqrt{(2n)!}} \right)^2=\frac{n!^2}{(2n)!}=\frac{1}{2n \choose n}$$

And

$${2n \choose n} \underset{n\to\infty}{\sim} \frac{4^n}{\sqrt{n\pi}}$$

So by ratio test, your series is convergent.

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  • $\begingroup$ What the ? Did you just approximately limit an ncr ? $\endgroup$ – Senishoshitsu Dec 21 '13 at 22:03
  • $\begingroup$ If $u_n \sim v_n$, convergence of $\sum u_n$ is equivalent to convergence of $\sum v_n$. Here it's a bit easier with $v_n=\sqrt{n\pi}/4^n$, since it's almost trivial to prove $v_{n+1}/v_{n} \to 1/4$. $\endgroup$ – Jean-Claude Arbaut Dec 21 '13 at 22:32
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Clearly $$ \frac{a_{n+1}}{a_n} = \frac{n+1}{\sqrt{(2n+1)(2n+2)}}\to \frac{1}{2}. $$ where $a_n=\dfrac{n!}{\sqrt{(2n)!}}$, and using the ratio test we get that the series converges absolutely.

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  • $\begingroup$ @AntonioVargas You are right! I corrected it! $\endgroup$ – Yiorgos S. Smyrlis Dec 21 '13 at 21:57
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There is another way to approach this problem using Stirling approximation for (n!).
http://en.wikipedia.org/wiki/Stirling_approximation
I just give this solution ignoring if you are or not allowed to use it. If you apply the second simplest approximation by Stirling
$$n! = \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n $$ it is easy, using the ratio test, that the limit is smaller than unity.
What is more interesting is that, using the previous formula, the summation (from $n=1$ to $\infty$) is then approximated by $$\pi^{1/4} \text{PolyLog}\left[-\frac14, \frac12\right] \approx 1.52783...$$ while the exact value (not using at all any approximation for $n!$) leads to $1.58955...$

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    $\begingroup$ If you don't use LaTeX in your answers there's a fair chance not many will take the time to try to decypher what's written there... $\endgroup$ – DonAntonio Dec 22 '13 at 4:51
  • $\begingroup$ @DonAntonio.I know that and I am so sorry of it. I think I already explained my problem to you and its becoming worse everyday. Whan my wife is close to my desk, she is able to make a very few things for me. Fortunately, from time to time, some users nicely edit my answers. Again, I can only apologize. Cheers. $\endgroup$ – Claude Leibovici Dec 22 '13 at 4:59
  • $\begingroup$ Oh, no need to apologize, @Claude. You owe me, or anyone else, no explanations at all. I don't remember what your problem is, and I was just pointing out about the ease to read stuff here. $\endgroup$ – DonAntonio Dec 22 '13 at 5:01
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    $\begingroup$ @DonAntonio. I thought I told you that I am "almost" (> 95%) blind, so I do not see what I am typing and evrything beside pure ASCII is very difficult to me. Moreover, ages don't make things easier. $\endgroup$ – Claude Leibovici Dec 22 '13 at 5:05

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