Proof convergence of $\sum\limits_{n = 1}^\infty\frac {n!} {n^n}$

Question

To prove the convergence of $\sum\limits_{n = 1}^\infty\frac {n!} {n^n}$ I used that $\lim\limits_{n\to\infty}|\frac {a_{n+1}} {a_n}|$ has to be $<1$:

$$\lim\limits_{n\to\infty}|\frac {a_{n+1}} {a_n}|$$

$$=\lim\limits_{n\to\infty}|\frac {\frac {(n+1)!} {(n+1)^{n+1}}} {\frac {n!} {n^n}}|$$

$$=\lim\limits_{n\to\infty}|\frac {(n+1)!\cdot n^n} {n!\cdot (n+1)^{n+1}}|$$

$$=\lim\limits_{n\to\infty}|\frac {(n+1)\cdot n^n} {(n+1)^{n+1}}|$$

$$=\lim\limits_{n\to\infty}|\frac {n^n} {(n+1)^n}|$$

$$=\lim\limits_{n\to\infty}|{(\frac {n} {n+1})}^n|=\frac 1 e$$

Okay... problem solved after thankful advice

You didn't apply $a_{n+1}$ correctly, it should be $(n+1)!/(n+1)^{n+1}$ — Matthew Cassell, Jan 11 '17 at 08:24
@Mattos I'm going to correct it, but the limit is still $0$ isn't it? — fatalError, Jan 11 '17 at 08:30
@GuyFabrice Wait I failed... I meant to say it remains $1$ why does it become 0 now? — fatalError, Jan 11 '17 at 08:37
The limit in the last step is in fact $\frac{1}{e}$. The problem is that as the exponent gets larger, the base also gets closer to $1$. — user361424, Jan 11 '17 at 09:45

Fernando Revilla · Answer 1 · 2017-01-11T08:46:27.707

1

We can also use the Comparison Test: $$0\le \frac{n!}{n^n}=\frac{1\cdot 2\cdot 3\cdot\ldots\cdot n}{n\cdot n\cdot n\cdot\ldots\cdot n}\le \frac{1\cdot 2\cdot n\cdot\ldots \cdot n}{n\cdot n\cdot n\cdot\ldots\cdot n}=\frac{2}{n^2},$$ and $\displaystyle \sum_{n=1}^{+\infty}\frac{2}{n^2}$is convergent, so the given series is convergent.

edited Jan 11 '17 at 08:46

answered Jan 11 '17 at 08:40

Fernando Revilla

3,537

score 0 · Answer 2 · answered Jan 11 '17 at 08:30

Your mistake happens at:

$$\frac{a_{n+1}}{a_n} = \frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}}\neq \frac {\frac {(n+1)!} {n^{n+1}}} {\frac {n!} {n^n}}$$

Continuing, you should get $$\frac{a_{n+1}}{a_n}=\frac{(n+1)!n^n}{n! (n+1)^{n+1}} = \frac{(n+1)\cdot n! \cdot n^n}{n!\cdot (n+1)\cdot (n+1)^n}$$

and you should be able to conclude from here.

Proof convergence of $\sum\limits_{n = 1}^\infty\frac {n!} {n^n}$

2 Answers2