# Convergence of $\sum_{n=1}^\infty\frac {n^{n}}{e^nn!}$

Check the convergence of: $\displaystyle\sum_{n=1}^\infty\frac {n^{n}}{e^nn!}$

Using the root test I get: $\displaystyle\lim_{n \to\infty} \dfrac {n}{e\sqrt[n]{n!}}$ now I'm left with showing that $n > \sqrt[n]{n!} \ \ \forall n$, can I just raise it to the power of $n$ like so: $\ n^n>n!$ ?

Alternatively, using limit arithmetic: $\displaystyle\lim_{n \to\infty} \dfrac {n}{e\sqrt[n]{n!}}=\displaystyle\lim_{n \to\infty} \dfrac {1}{\large\frac e n \sqrt[n]{\frac {n!}{n^n}}}>1$ (that's not very persuasive I know) so it diverges.

Edit: Root test won't work.

Note: Stirling, Taylor or integration are not allowed.

• This looks like a reversal of Stirling's approximation; as such, I believe this works out to a sum over an infinite number of positive, non-zero constants that do not approach zero... – abiessu Apr 22 '14 at 16:37
• No, that doesn't work. Your limit has value $1$, unfortunately. See this. – David Mitra Apr 22 '14 at 16:41
• So the root test doesn't work... – GinKin Apr 22 '14 at 16:43
• Stirling Approx not allowed again? – Santosh Linkha Apr 22 '14 at 16:46
• No it's not allowed @SantoshLinkha – GinKin Apr 22 '14 at 16:47

Let $$u_n=\frac{n^n}{e^nn!}$$ so $$\frac{u_{n+1}}{u_n}=\frac1e \left(1+\frac1n\right)^n=\frac1e\exp\left(n\left(\frac1n-\frac1{2n^2}+o\left(\frac1{n^2}\right)\right)\right)=\exp\left(-\frac1{2n}+o\left(\frac1{n}\right)\right)\\=1-\frac1{2n}+o\left(\frac1{n}\right)$$ so using the Raabe-Duhamel's rule the series is convergent.

• Thanks but I'm not allowed to use Taylor or any approximation. We don't cover this rule either. – GinKin Apr 22 '14 at 17:04
• Sami: I'm sorry about the reversal... I was much more careful today. I hope you're not too upset with me? – Namaste Apr 23 '14 at 15:47

An idea: put

$$a_n:=\frac{n!}{n^n}\implies \frac{a_{n+1}}{a_n}=\frac{(n+1)!n^n}{(n+1)^{n+1}n!}=\frac1{\left(1+\frac1n\right)^n}\xrightarrow[n\to\infty]{}\frac1e<1$$

Thus, the series $\;\sum a_n\;$ converges by the quotient text, and from here

$$a_n=\frac{n!}{n^n}\xrightarrow[n\to\infty]{}0$$

and from here, for some $\;n\;$ on, we get that

$$n!\le n^n\implies\sqrt[n]{n!}\le n$$

• Thanks. The root test doesn't work here though. – GinKin Apr 23 '14 at 7:52
• Nop, it doesn't, but this complete's the root test applied to tthe original series. – DonAntonio Apr 23 '14 at 13:19