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

Question

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. — ajotatxe, Jan 20 '16 at 13:13
No problem. I just edited it so people wouldn't get stuck over something as simple as a misplaced index. — zz20s, Jan 20 '16 at 13:18
That use of l'Hopital would be OK for a fixed exponent, but this one goes to infinity. — GEdgar, Jan 20 '16 at 14:15
Duplicate of Is $\sum\limits_{n=1}^{\infty}\frac{n!}{n^n}$ convergent? — metamorphy, Mar 29 '23 at 04:01

score 17 · Answer 1 · answered Jan 20 '16 at 13:17

17

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.

answered Jan 20 '16 at 13:17

ajotatxe

65,084

Do you "remember" these types of inequalities from experience? – Vizuna Jan 20 '16 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. – Vizuna Jan 20 '16 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 $1$. – ajotatxe Jan 20 '16 at 14:11

score 10 · Accepted Answer · answered Jan 20 '16 at 13:23

10

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.

answered Jan 20 '16 at 13:23

SchrodingersCat

24,472

(+1) I was working on mine when you posted yours. I will delete mine. – robjohn Jan 20 '16 at 13:30
Thanks for the answer. I am sorta confused about how did you get $\frac{1}{e}$ in the last step? – Jay Wang Jan 20 '16 at 13:40
By definition. That's a definition. $e=\lim_{x\to \infty} (1+\frac{1}{x})^x$ – SchrodingersCat Jan 20 '16 at 13:41

score 9 · Answer 3 · answered Jan 20 '16 at 13:15

9

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

answered Jan 20 '16 at 13:15

egreg

238,574

score 4 · Answer 4 · answered Jan 20 '16 at 13:26

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 $z\,e^{1-z}$.

Tom-Tom · Answer 5 · 2016-01-20T16:08:11.290

2

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 $$\ln k!\approx k\ln k-k+\frac12\ln k+\frac12\ln(2\pi)+O(k^{-1})$$

edited Jan 20 '16 at 16:08

answered Jan 20 '16 at 13:14

Tom-Tom

6,867

Mathematica gives the sum of the series as approximately 1.87985386217525. – Tom-Tom Jan 20 '16 at 13:17
Someone doesn't like Mr. Stirling here... – Tom-Tom Jan 20 '16 at 13:44
1

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.) – Travis Willse Jan 20 '16 at 14:21
@Travis. Thanks for the comment. I have edited my answer. – Tom-Tom Jan 20 '16 at 16:09

score 1 · Answer 6 · answered Jan 20 '16 at 13:17

Notice:

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

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

6 Answers6

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