# Is $\sum\limits_{n=1}^{\infty}\frac{n!}{n^n}$ convergent?

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.

• Hint: ratio test – Willie Wong Nov 11 '11 at 14:06
• physicsforums.com/showthread.php?t=357784 – Martin Sleziak Nov 11 '11 at 14:14
• 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

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.
• 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
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 $k$.
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 $n \geq 2$).
• 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