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

Here's another series I haven't been able to crack only with the comparison test (which is required because it is an exercise given before having explained the other tests regarding series). It seems to me that it converges, and it occured to me that I could use the fact that $$n^n \leq (n!)^2$$ but this hasn't helped me much.

I know Stirling's formula is inviting but I don't think it would be fair either. Sorry guys, I know I'm being pedantic.

  • $\begingroup$ At the risk of being careless in the analysis, I find that problems of this nature are usually solved by first applying Stirling's formula to the factorial. $\endgroup$ – JessicaK Apr 7 '14 at 11:06
  • $\begingroup$ Well, it may be possible, but this is sadly another thing which is much later in the book, so I don't think it would be fair to use it. $\endgroup$ – Gennaro Marco Devincenzis Apr 7 '14 at 11:09
  • $\begingroup$ Are you allowed to use the ratio test? $\endgroup$ – Jimmy R. Apr 7 '14 at 11:12
  • $\begingroup$ Nope. I suppose I could prove it before using it here, but again it is usually an "official" proof and I'd rather not use it. Besides, I've found that manipulating inequalities is the analyst's favourite hobby, so it may be instructive :P $\endgroup$ – Gennaro Marco Devincenzis Apr 7 '14 at 11:15

Use the Striling formula:

$$\frac{n^n}{(n!)^2} \sim \frac{n^n}{n! n^{n} \sqrt{2\pi n} e^{-n}} = O\left(\frac { e^{n}}{n! \sqrt{ n} }\right) = O\left( \frac { e^{n}}{n! }\right) $$the series of which converges.

Another way:

$$ (n!)^2 = \prod_{k=1}^n k(n-k+1) \ge \left(\frac{n+1}2\right)^{2n} \\ \frac{n^n}{(n!)^2} \le \left(\frac {4n}{(n+1)^2}\right)^{n}\le 2^{-n} $$ for $n$ big enough, because $$ x\to -x(n-x+1) $$is convex with a minimum when its derivative is $0$.

  • $\begingroup$ What is $$\prod_{k=1}^n k(n-k+1) \ge \prod_{k=1}^n k(n-k+1)$$ supposed to prove? $x\geq x$? Probably a typo somewhere here... $\endgroup$ – 5xum Apr 7 '14 at 11:17
  • $\begingroup$ yep. thanks. {}{} $\endgroup$ – mookid Apr 7 '14 at 11:18
  • $\begingroup$ $e^n/n!$. Did you just come up with $e^e$? =) $\endgroup$ – Guy Apr 7 '14 at 11:40
  • $\begingroup$ do not forget the $O(.)$ ;) $\endgroup$ – mookid Apr 7 '14 at 11:45

By the harmonic and geometric means inequality you have $$\sqrt[n]{n!}\geq \frac n{H_n}$$ where $H_n=\sum\limits_{k=1}^n\frac1k$.

Then, $$\frac{n^n}{(n!)^2}\leq\frac{n^n}{(n^n/H_n^n)^2}=\left(\frac{H_n^2}n\right)^n<\left(\frac{(1+\log n)^2}{n}\right)^n$$.

Now, define $$f(x)=\frac{(1+\log x)^2}{x}$$ and apply l'Hopital's rule to get $$\lim_{x\rightarrow\infty}f(x)=\lim_{x\rightarrow\infty}\frac2x(1+\log x)=0$$ so there exists some $n_0\in\Bbb N$ such that $$\frac{(1+\log n)^2}{n}<\frac12$$ so the series is convergent.


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