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

Question

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

score 2 · Answer 1 · edited May 01 '20 at 03:23

2

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 divergent.

edited May 01 '20 at 03:23

Ri-Li

9,038

answered Apr 22 '14 at 16:55

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? – amWhy Apr 23 '14 at 15:47
It seems something is wrong here, see the linked question – Vincent Nov 24 '19 at 10:30

score 2 · Answer 2 · answered Apr 22 '14 at 19:02

2

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

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

answered Apr 22 '14 at 19:02

DonAntonio

211,718
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Thanks. The root test doesn't work here though. – GinKin Apr 23 '14 at 07:52
Nop, it doesn't, but this complete's the root test applied to tthe original series. – DonAntonio Apr 23 '14 at 13:19
So basically you proved here $n! \leq n^n$? – Ri-Li May 01 '20 at 03:16
@Ri-Li Yes, of course. That is exactly the very last inequality. – DonAntonio May 01 '20 at 07:28

score 1 · Answer 3 · answered May 01 '20 at 03:48

Let us give another proof:

Lemma: $\sum a_n$, $\sum b_n$ are series of positive terms. If we have $\frac{a_{n+1}}{a_n}\leq \frac{b_{n+1}}{b_n}$ then if $\sum b_n$ is convergent so is $\sum a_n$.

proof: $\frac{a_{n+1}}{a_n}\leq \frac{b_{n+1}}{b_n} \Rightarrow \frac{a_{n+1}}{b_{n+1}}\leq \frac{a_n}{b_n}$.

Consider $c_n=\frac{a_n}{b_n}$ then $\frac{a_{n+1}}{b_{n+1}}\leq \frac{a_n}{b_n}\Rightarrow c_{n+1}\leq c_n\leq c_1\Rightarrow \frac{a_n}{b_n}\leq c_1 \Rightarrow a_n \leq c_1 b_n$.

Now $\sum b_n$ is convergent so is $\sum a_n$[by comparison test].

Now let us assume $a_n=\displaystyle\sum_{n=1}^\infty\frac {n^{n}}{e^nn!}$

Then $\frac{a_{n+1}}{a_n}=\displaystyle \dfrac {(n+1)^n}{en^n}=\displaystyle \frac {\left(1+\dfrac {1}{n}\right)^n}{e}>\displaystyle \frac {\left(1+\dfrac {1}{n}\right)^n}{\left(1+\dfrac {1}{n}\right)^{n+1}}=\displaystyle \frac {1}{\left(1+\dfrac {1}{n}\right)}=\dfrac {n}{n+1}=\dfrac {\frac{1}{n+1}}{\frac 1n}$.

Applying last lemma if $\sum a_n$ is convergent then $\sum \frac 1n$ is convergent.

Contradiction!!

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

3 Answers3