# Does $\sum_{k=1}^{\infty}\frac{k!}{k^k}$ converge?

I have tried using ratio test: $$P =\lim_{k\rightarrow\infty}\left|\frac{(k+1)!}{(k+1)^{k+1}}\cdot\frac{k^k}{k!}\right|$$ $$P=\lim_{k\rightarrow\infty}\left|\frac{(k+1)\cdot k^k}{(k+1)^{k+1}}\right|$$ $$P=\lim_{k\rightarrow\infty}\left|\frac{k^{k+1}+ k^k}{(k+1)^{k+1}}\right|$$

In the final expression, the highest degrees in the denominator and numerator are both $(k+1)$. So according to the L'Hospital's Rule, the limit would goes to $1$.

$$P=1$$

Thus the test failed.

Any suggestions on how to test the convergence of this series?

• You use $k$ in the expressions, but $n$ under sum and limit symbols. Jan 20, 2016 at 13:13
• Sum on $n$ and index is $k$..??? Jan 20, 2016 at 13:14
• I went ahead and edited the post. Jan 20, 2016 at 13:15
• Sorry... edited Jan 20, 2016 at 13:16
• That use of l'Hopital would be OK for a fixed exponent, but this one goes to infinity. Jan 20, 2016 at 14:15

For $n\ge 3$ we have $$\frac{n!}{n^n}\le\frac{1\cdot 2}{n^2}$$ Since the series $\sum n^{-2}$ converges, your series converges, too.

• Do you "remember" these types of inequalities from experience? Jan 20, 2016 at 13:50
• I guess by writing out the first few terms we could easily see the same inequality. Just curious about your thought process though. Jan 20, 2016 at 13:54
• I simply wrote an example and saw that the two first factors were $1/n$ and $2/n$, being the remainding factors lesser than $1$. Jan 20, 2016 at 14:11

You should proceed as follows:

$$P =\lim_{k\rightarrow\infty}\left|\frac{(k+1)!}{(k+1)^{k+1}}\cdot\frac{k^k}{k!}\right|$$ or $$P=\lim_{k\rightarrow\infty}\left|\frac{(k+1)\cdot k^k}{(k+1)^{k+1}}\right|$$ or $$P=\lim_{k\rightarrow\infty}\left|\frac{k^k}{(k+1)^{k}}\right|$$ or $$P=\frac{1}{\lim_\limits{k\rightarrow\infty}\left(1+\frac{1}{k}\right)^k}$$ or $$P=\frac{1}{e}$$ by definition.

• (+1) I was working on mine when you posted yours. I will delete mine.
– robjohn
Jan 20, 2016 at 13:30
• Thanks for the answer. I am sorta confused about how did you get $\frac{1}{e}$ in the last step? Jan 20, 2016 at 13:40
• By definition. That's a definition. $e=\lim_{x\to \infty} (1+\frac{1}{x})^x$ Jan 20, 2016 at 13:41

Note that $$\frac{(k+1)k^k}{(k+1)^{k+1}}=\frac{k^k}{(k+1)^k}= \left(\left(1+\frac{1}{k}\right)^{\!k}\right)^{-1}$$

We have:

$$\sum_{k=1}^{N}\frac{k!}{k^k} = \int_{0}^{+\infty}e^{-x}\sum_{k=1}^{N}\frac{x^k}{k^k}\,dx = \int_{0}^{+\infty}\sum_{k=1}^{N}k e^{-kz}z^k\,dz\tag{1}$$ hence the series is convergent since the function $$f(z)=\frac{z e^z}{(e^z-z)^2}=\sum_{k\geq 1}ke^{-kz}z^k\tag{2}$$ is integrable over $\mathbb{R}^+$, as a positive function bounded by $z\,e^{1-z}$.

Stirling's approximation gives $$k!\approx k^k\mathrm e^{-k}\sqrt{2\pi k}\;\left(1+O\Big(\frac1k\Big)\right)$$ therefore the generic term is equivalent to $$\frac{k!}{k^k}\sim\mathrm e^{-k}\sqrt{2\pi k}$$ and the series converges.

EDIT I have added a precision concerning Stirling approximation. Usually Stirling approximation is given in logarithmic form $$\ln k!\approx k\ln k-k+\frac12\ln k+\frac12\ln(2\pi)+O(k^{-1})$$

• Mathematica gives the sum of the series as approximately 1.87985386217525. Jan 20, 2016 at 13:17
• Someone doesn't like Mr. Stirling here... Jan 20, 2016 at 13:44
• I did not downvote, but perhaps the objection is that, as an approximation, one needs appropriate error bounds to establish convergence using this technique, and those are not cited. (In this case, it's not too hard to analyze one of the common bounds given for Stirling and hence draw the same conclusion.) Jan 20, 2016 at 14:21
• @Travis. Thanks for the comment. I have edited my answer. Jan 20, 2016 at 16:09

Notice:

$$\left|\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!}\right|=\left|\left(\frac{n}{n+1}\right)^n\right|=$$ $$\left|\left(\frac{1}{1+\frac{1}{n}}\right)^n\right|=\left|\frac{1}{\left(1+\frac{1}{n}\right)^n}\right|=$$ $$\frac{1}{\left|\left(1+\frac{1}{n}\right)^n\right|}=\frac{1}{\left|1+\frac{1}{n}\right|^n}$$