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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 ?
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
$$ \begin{align} \frac{\frac{1}{\sqrt{n}}\sqrt{n+1}}{\frac{\sqrt{2}}{\sqrt{n}}\sqrt{2n+1}} &= \frac{\sqrt{1+\frac{1}{n}}}{\sqrt{2}\sqrt{2 + \frac{1}{n}}} \\ &\overset{n\to\infty}{\longrightarrow} \frac{\sqrt{1+0}}{\sqrt{2}\sqrt{2 + 0}} \\ &= \frac{1}{2}. \end{align} $$
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.
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.
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 $1.58955...$
