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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.

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@dtldarek That question does not answer this one. That fact doesn't imply that this converges or diverges necessarily. –  Andrew Salmon Mar 6 '13 at 7:17
    
i think i get the answer, using the ratio test –  AAA Mar 6 '13 at 7:22
    
@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 6 '13 at 9:02
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4 Answers

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

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.

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+1. Way easier! @myself: Why do I always think of Stirling when I see a factorial? –  user17762 Mar 6 '13 at 7:24
    
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 6 '13 at 7:24
    
@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 6 '13 at 7:27
    
@HaraldHanche-Olsen I just added a remark, that convergence and the limit need not even be known. –  Hagen von Eitzen Mar 6 '13 at 7:31
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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

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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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damnet, beat me to it –  Ethan Mar 6 '13 at 7:19
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$$\frac{n!}{n^n} = \frac{1}{n} . \frac{2}{n} \dots \frac{n}{n} \leq \frac{1}{n} .\frac{2}{n} . 1 \times 1 \dots \times 1 = \frac{2}{n^2}. $$

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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: mathforum.org/library/drmath/view/53380.html –  user1354557 Mar 6 '13 at 21:01
    
@user1354557 To finish, the integral test reveals convergence of $\sum 2/n^2$. –  QuantumDot Feb 9 at 8:49
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