Testing a series for convergence - is my solution correct?

**thomasdotnet** · September 6th 2009, 12:17 PM

Hello everyone,

I tested the following series for convergence, but I'm not sure if all the steps I took are strictly legal. Thus I would like to ask you to take a look at them.

$\sum_{n=1}^{+\infty}{\frac{\left(n!\right)^2}{n^n} }$

I use the following fact to construct a series that is less than the one above for a comparison test.

$n!>\sqrt{2\pi n}\left(\frac{n}{e}\right)^n$

$\sum_{n=1}^{+\infty} {\frac{\left(\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\right)^2}{n^n}}$

Now I conduct a root test for this series to test it for convergence.

$\lim\limits_{n\rightarrow+\infty} {\frac{\sqrt[n]{2\pi n\left(\frac{n}{e}\right)^{2n}}}{\sqrt[n]{n^n}}} = \lim\limits_{n\rightarrow+\infty} {\frac{\sqrt[n]{2\pi n}\left(\frac{n}{e}\right)^2}{n}} = \lim\limits_{n\rightarrow+\infty} {\sqrt[n]{2\pi n}\frac{n}{e^2}} = \lim\limits_{n\rightarrow+\infty} {e^{\frac{1}{n}\log{\left(2\pi n\right)}}\frac{n}{e^2}} $

At this point the exponent of the exponential function is the problem, which I solve by using l'Hôpital's rule. (I'm not exactly sure if I am allowed to do this here.)

$\lim\limits_{n\rightarrow+\infty}{\frac{\log{\left (2\pi n\right)}}{n}} = \lim\limits_{n\rightarrow+\infty} {\frac{2\pi}{2\pi n}}=\lim\limits_{n\rightarrow+\infty} {\frac{1}{n}}$

Now I move this newly obtained exponent back into the original limit calculation.

$\lim\limits_{n\rightarrow+\infty} {e^{\frac{1}{n}}\frac{n}{e^2}}=+\infty$

Since the smaller series diverges, also the greater one has to diverge.

I would appreciate it, if someone could tell me if I am right.

**Enrique2** · September 6th 2009, 01:33 PM

The l'Hopital rule can be applied to obtain the limit when $x\to\infty$ , hence if the limit exist the discrete limit when $n\to\infty$ also exists and agree. So this step is correct.

Also calculating the limit of the general term and observing that is infinty will give you the divergence of the series.

**Plato** · September 6th 2009, 01:43 PM

$\sqrt[n]{{\frac{{\left( {n!} \right)^2 }}{{n^n }}}} = \sqrt[n]{{\frac{{\left( {n!} \right)}}{{n^n }}}}\sqrt[n]{{n!}}$

$\left( {\sqrt[n]{{\frac{{\left( {n!} \right)}}{{n^n }}}}} \right) \to \frac{1}{e}$

**thomasdotnet** · September 7th 2009, 03:03 AM

Thank you very much for your replies!
I found out that l'Hôpital's rule is not to be used for sequences, except if the sequence can be transformed into a function $f(x)$ such that $f(n)=a_n$ , which is fortunately the case here.

Originally Posted by Plato

$\sqrt[n]{{\frac{{\left( {n!} \right)^2 }}{{n^n }}}} = \sqrt[n]{{\frac{{\left( {n!} \right)}}{{n^n }}}}\sqrt[n]{{n!}}$

$\left( {\sqrt[n]{{\frac{{\left( {n!} \right)}}{{n^n }}}}} \right) \to \frac{1}{e}$

Thanks for that hint Plato, but how I can I prove that $\sqrt[n]{{n!}}\rightarrow+\infty$ ?

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