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We have the following series

$$ \displaystyle \sum_{n=1}^\infty \dfrac{n!}{(2n)!} 2^n$$

I want to test this for convergence using the ratio test, but I don't know how I can get rid of those factorials or at least get an expression for which I can find the limit.

Is there another test you should apply?

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  • $\begingroup$ The ratio test is perfect for this job. Try it again! $\endgroup$ – user98602 Dec 15 '15 at 23:11
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The ratio test is the best test. The usual technique is to write $(n + 1)! = (n + 1) \cdot n!$; for example,

\begin{align*} \frac{\frac{n! 2^n}{(2n)!}}{\frac{(n + 1)! 2^{n + 1}}{(2(n + 1)!)}} &= \frac{n!}{(n + 1)!}\frac{2^n}{2^{n + 1}} \frac{(2(n + 1))!}{(2n)!} \end{align*} The first term simplifies as $1/(n + 1)$, for example; the others are similar.


Alternatively, convince yourself that $(2n)! \ge (n!)^2$ for large enough $n$, and then notice

$$\sum_{n = 1}^{\infty} \frac{2^n}{n!} = e^2 < \infty$$

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The ratio test should work just fine here:

$\dfrac{a_{n+1}}{a_n} = \dfrac{\dfrac{(n+1)!}{(2n+2)!} \cdot 2^{n+1}}{\dfrac{n!}{(2n)!} \cdot 2^n} = \dfrac{\dfrac{(n+1) \cdot n!}{(2n+2) \cdot (2n+1) \cdot (2n)!} \cdot 2 \cdot 2^{n}}{\dfrac{n!}{(2n)!} \cdot 2^n}$

Now, all you need to do is cancel stuff and take the limit as $n \to \infty$.

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