Prove that $\sum_{k=1}^\infty \frac{e^kk!}{k^k}$ does not converge.

Question

I am doing an exercise where I have to consider the radius and interval of convergence of a series of functions. I found one endpoint to be $e/2$ and I now need to consider whether or not the following sum diverges. $$\sum_{k=1}^\infty \frac{e^kk!}{k^k}$$

I have found from calculators that it diverges but I do not know how to prove this.

Cauchy's ratiotest is inconclusive, as it gives the value 1.

Any help?

Thanks for your comment but that's not been covered so there must be another way? — QuestionMaker, Jan 10 '17 at 19:03
Hey Dr. Graubner how would I calculate that limit? Should I substitute $e=(\frac{k+1}{k})^k$? I tried that but it only seems to make things messy. @Dr.SonnhardGraubner — QuestionMaker, Jan 10 '17 at 19:07
i would compute $$\frac{a_{k+1}}{a_k}$$ for the sequence $$a_k=\frac{e^k k!}{k^k}$$ — Dr. Sonnhard Graubner, Jan 10 '17 at 19:09
That fraction equals $e (\frac{k}{k+1}) ^k$, of which the limit as $k$ goes to infinity is equal to $1$, so I can't apply a ratio test there, can I? @Dr.SonnhardGraubner — QuestionMaker, Jan 10 '17 at 19:12
$\frac{a_{k+1}}{a_k}=\frac{e^{k+1}}{e^k}\frac{(k+1)!}{k!}\frac{k^k}{(k+1)^{k+1}}=e(k+1)\frac{k^k}{(k+1)^{k+1}}=e\frac{k^k}{(k+1)^{k}}=e(\frac{k}{k+1})$ — QuestionMaker, Jan 10 '17 at 19:19
@QuestionMaker, We can proceed from Dr. Sonnhard Graubner's comment by utilizing the inequality that $(1 + \frac{1}{k})^k < e$. This can be proved in various ways, for instance, you can show that $(1 + \frac{1}{k})^{k}$ strictly increases using Bernoulli's inequality, or you can apply the binomial theorem to expand $(1+\frac{1}{k})^k$ and compare the resulting sum with $e = \sum_{j=0}^{\infty} \frac{1}{j!}$. — Sangchul Lee, Jan 10 '17 at 19:20
Thanks guys, I got it. I used indeed Bernoulli to prove that the strictly increasing which lead to Dr. Graubner's fraction to be more or equal to 1. — QuestionMaker, Jan 10 '17 at 20:07

score 6 · Accepted Answer · answered Jan 10 '17 at 19:04

6

If you expand $e^k$ you will find a term $\cfrac {k^k}{k!}$ and all other terms are positive.

answered Jan 10 '17 at 19:04

Mark Bennet

100,194

score 2 · Answer 2 · answered Jan 10 '17 at 19:02

2

Stirling's approximation tells us $n! = \sqrt{2 \pi n}\frac{n^n}{e^n} (1+ O(1/n))$.

Now, plug this into the summand and see what it behaves like.

answered Jan 10 '17 at 19:02

Batman

19,390

Prove that $\sum_{k=1}^\infty \frac{e^kk!}{k^k}$ does not converge.

2 Answers2