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I can clearly see that $\dfrac{n!}{n^n}\to 0$ when $n\to\infty$. But how do I know if the sum $$\sum_{n=1}^{\infty}\frac{n!}{n^n}$$ is convergent or not? I know this might be basic, but thank you if anyone can help me.

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Hint: ratio test –  Willie Wong Nov 11 '11 at 14:06
    
    
See also this answer: math.stackexchange.com/questions/48897/… –  Martin Sleziak Nov 11 '11 at 14:21
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We can also use the comparison test: Write each term as $$\frac{n!}{n^n}=\left(\frac{n}{n}\right)\left(\frac{n-1}{n}\right)\left(\frac{n‌​-2}{n}\right)\cdots \left(\frac{2}{n}\right)\left(\frac{1}{n}\right).$$ All of these parts are $\leq 1$, and half of the terms are $\leq \frac{1}{2}$ so we see that $$\frac{n!}{n^n}\leq \frac{1}{2^{(n-1)/2}}.$$ These last terms form a geometric series. –  Eric Naslund Nov 11 '11 at 16:42
    
@Eric: This is simple and requires no reference to Stirling. Why not make it into an answer? –  robjohn Nov 11 '11 at 17:33
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3 Answers 3

up vote 7 down vote accepted

This is normally done by using one of the series convergence tests. In your example, ratio test appears applicable, which looks at the limit of ratio of subsequent terms in the series: $R=\lim_{n\to\infty} |a_{n+1}/a_n|$. If you can show that $R<1$, then the original series $\sum_{n=1}^{\infty}a_n$ converges absolutely. For your specific series $$ \frac{a_{n+1}}{a_n}=\frac{(n+1)!}{(n+1)^{n+1}}/\frac{n!}{n^n}=\frac{n^n}{(n+1)^n}. $$ When $n$ is large $(n+1)^n=\exp(n\ln(n+1))=\exp(n(\ln n+1/n+O(n^{-2})))\approx en^n$, from which it can be concluded that $$ \lim_{n\to\infty}\frac{n^n}{(n+1)^n}=\frac{1}{e}<1. $$ Therefore, your series converges absolutely.

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This is a basic Calc 2 question. Students probably don't know things such as $O(n^{-2})$. On the other hand, they have probably seen that $\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e$. Or, even if they have not, they should have seen that they can take the $\ln$ of such a thing to find the limit. Either way, they can figure out that is $e$, so they have a rule saying the limit of the reciprocal is the reciprocal of the limit. We must answer questions in ways the askers can understand. Your answer is better than Gortaur's giving Stirling's formula at least. –  Graphth Nov 11 '11 at 14:36
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@Graphth: You are correct, I am slightly confused by the level of complexity expected here. However, I'd rather push student to think in terms of the truncated series and basic algebraic transformations, than expect the student to memorize certain specific limits without proper understanding of where they come from. –  Aleksey Pichugin Nov 11 '11 at 14:47
    
In some sense, I see what you are saying. I see people using these truncated series all the time. They are very useful. It would have been nice if I learned about them earlier. On the other hand, the limit I gave is one of the most important limits in math, so it would be good if they memorized that one. –  Graphth Nov 11 '11 at 14:54
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You can also use Stirling's_formula if you already know it to obtain that $$ \frac{n!}{n^n}\sim \sqrt{2\pi n}\mathrm \cdot e^{-n} $$ and then note that series $\sum\limits_{n=1}^\infty n^k \cdot\mathrm e^{-n}$ converge for any finite $k$.

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Hint: $n!/n^n = (n/n) ((n-1)/n) ((n-2)/n) \cdots (2/n) (1/n)$. Each term in the product is less than or equal to one. So the whole thing is at most $2/n^2$ (for $n \geq 2$).

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I know you are making the right argument, but the wording of the last sentence is off logically: "Each term in the product is less than or equal to one. So the whole thing is at most $2/n^2.$" This certainly isn't true for all products! –  Eric Naslund Nov 11 '11 at 16:39
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(Silly comment: products do not have terms but factors) –  Mariano Suárez-Alvarez Nov 11 '11 at 17:18
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