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

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 ?

• 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. Dec 21, 2013 at 21:34
• @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. Dec 21, 2013 at 21:38

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}

• Woah community wiki, thanks for clearing that out. Dec 21, 2013 at 22:02
• You're welcome :) Dec 22, 2013 at 0:51

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.

• What the ? Did you just approximately limit an ncr ? Dec 21, 2013 at 22:03
• 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 $v_{n+1}/v_{n} \to 1/4$. Dec 21, 2013 at 22:32

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.

• @AntonioVargas You are right! I corrected it! Dec 21, 2013 at 21:57

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 $1.58955...$

• 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... Dec 22, 2013 at 4:51
• @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. Dec 22, 2013 at 4:59
• 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. Dec 22, 2013 at 5:01
• @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. Dec 22, 2013 at 5:05