Find the convergence of the following sequence: $$\sum_{n=1}^{\infty}\left(\dfrac{n!}{(2n)!}\right)^{\frac{1}{n}}$$ I tried to use the ratio test. It wasn't useful. Also I used the logharitmic test (I'm not sure this is the standard name for it): $\lim_{n\to\infty}\dfrac{\ln\frac{1}{u_n}}{\ln n}=l$. If $l>1$, $\sum_{n=1}^{\infty}u_n$ converges. It didn't work to well. Thanks for your help!
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2$\begingroup$ are you aware of strilings formula? $\endgroup$– tiredCommented Nov 22, 2016 at 10:32
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2$\begingroup$ @tired: it seems that you are not the only one to know about this Strilings. show.docjava.com/cpe471/lectures/slides.htm/error_dt/… $\endgroup$– user65203Commented Nov 22, 2016 at 10:36
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$\begingroup$ No, "logharitmic" is not standard, though Google finds 885 instances. $\endgroup$– user65203Commented Nov 22, 2016 at 11:48
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$\begingroup$ @YvesDaoust - you may have confused him with that last comment. (Curiously, Google only found 880 instances for me - I feel cheated...). To explain: It is spelled "logarithmic", and yes that is the standard name for the test. $\endgroup$– Paul SinclairCommented Nov 22, 2016 at 17:52
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2 Answers
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$$\left(\frac{n!}{(2n)!}\right)^{1/n}=\left(\frac1{(n+1)(n+2)\cdots(2n)}\right)^{1/n}>\left(\frac1{(2n)(2n)\cdots(2n)}\right)^{1/n}=\frac1{2n}$$
hence the series diverges.
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Using Stirling's approximation, you get $$a_n=\left(\dfrac{n!}{(2n)!}\right)^{\frac{1}{n}} \sim \left( \frac{\sqrt{2 \pi n}}{\sqrt{4 \pi n}}\left(\frac{n}{e}\right)^n \left(\frac{e}{2n}\right)^{2n}\right)^{1/n}=\left(\frac{1}{\sqrt{2}}\right)^{1/n} \frac{e}{2n}$$
Hence the series diverges.
answered Nov 22, 2016 at 10:38