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Does $\sum \dfrac{n!}{n^n}$ converge? Intuitively I think it converges but I am not sure what test can be used to show that.

Gigili
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AAA
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    @dtldarek That question does not answer this one. That fact doesn't imply that this converges or diverges necessarily. – A.S Mar 06 '13 at 07:17
  • i think i get the answer, using the ratio test – AAA Mar 06 '13 at 07:22
  • @AndrewSalmon Yep, but the posts there include the answer to this one (e.g. using convergence of the series to show that it converges to zero). On the other hand, definitely not a duplicate (I have no idea what I was thinking). – dtldarek Mar 06 '13 at 09:02

4 Answers4

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$$\left|\frac{a_{n+1}}{a_n}\right|=\frac{(n+1)!\cdot n^n}{(n+1)^{n+1}\cdot n!} =\frac{n^n}{(n+1)^n}=\frac1{\left(1+\frac1n\right)^n}\to\frac1e$$

Remark: While $\lim_{n\to \infty} \left(1+\frac1n\right)^n$ it is often used as a definition of $e$, we don't even need convergence here. It suffices to have $\left(1+\frac1n\right)^n\ge 2$ (for $n\ge 1$), which follows from the Bernoulli inequality or simply from the first two summands of the binomial expansion.

Hagen von Eitzen
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  • +1. Way easier! @myself: Why do I always think of Stirling when I see a factorial? –  Mar 06 '13 at 07:24
  • I much prefer this solution because the limit of $(1+x/n)^n$ as $n\to\infty$ is quite well known, and easier to prove than Stirling's formula. – Harald Hanche-Olsen Mar 06 '13 at 07:24
  • @Marvis I was just about to make a comment to that effect - This is at least the third question in the last 24 hours involving factoirials, getting a (nice and correct) answer involving Stirling but allowing a short direct and elementary approach :) – Hagen von Eitzen Mar 06 '13 at 07:27
  • @HaraldHanche-Olsen I just added a remark, that convergence and the limit need not even be known. – Hagen von Eitzen Mar 06 '13 at 07:31
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If you don't know stirlings formula, by the inequality of arithmetic and geometric means we have, $$(\prod_{k=1}^n k)^{\frac{1}{n}}\leq\frac{1}{n}\sum_{k=1}^n k=\frac{n+1}{2}$$ $$\text{ So we have}$$ $$(n!)^\frac{1}{n}\leq\frac{n+1}{2}$$ $$\frac{n!}{(n+1)^n}\leq\frac{1}{2^n}$$ $$\frac{(n+1)!}{(n+1)^n}\leq\frac{(n+1)}{2^n}$$ $$\sum_{n=0}^\infty\frac{(n+1)!}{(n+1)^n}\leq\sum_{n=0}^\infty\frac{(n+1)}{2^n}=4$$ So, $\sum_{n=0}^\infty\frac{(n+1)!}{(n+1)^n}$, converges

Ethan Splaver
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From Stirling's formula, we have $$n! \sim \sqrt{2 \pi n} \left(\dfrac{n}e \right)^n$$ Hence, $$\dfrac{n!}{n^n} \sim \dfrac{\sqrt{2 \pi n}}{e^n} \implies \left(\dfrac{n!}{n^n} \right)^{1/n} \sim \dfrac{\sqrt[2n]{2 \pi n}}{e} \to \dfrac1e < 1$$ Hence, $\displaystyle \sum_{n}\dfrac{n!}{n^n}$ converges.

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$$\frac{n!}{n^n} = \frac{1}{n} . \frac{2}{n} \dots \frac{n}{n} \leq \frac{1}{n} .\frac{2}{n} . 1 \times 1 \dots \times 1 = \frac{2}{n^2}. $$

user10676
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  • +1. Although this is only a partial solution, I like this approach the most because it's very simple and it doesn't rely on other theorems. However, it isn't intuitively obvious that 2/n^2 converges, so this needs to be proved as well. My favorite method is explained here: http://mathforum.org/library/drmath/view/53380.html – user1354557 Mar 06 '13 at 21:01
  • @user1354557 To finish, the integral test reveals convergence of $\sum 2/n^2$. – QuantumDot Feb 09 '14 at 08:49