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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?

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    Use Strilings' approximation of $k!$ – 2017-01-10
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    prove that $$\lim_{k \to \infty}\frac{e^k k!}{k^k}=\infty$$ – 2017-01-10
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    Thanks for your comment but that's not been covered so there must be another way? – 2017-01-10
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    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 – 2017-01-10
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    i would compute $$\frac{a_{k+1}}{a_k}$$ for the sequence $$a_k=\frac{e^k k!}{k^k}$$ – 2017-01-10
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    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 – 2017-01-10
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    can you post your steps please? – 2017-01-10
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    $\frac{a_{k+1}}{a_k}=\frac{e^{k+1}}{e^k}\frac{(k+1)!}{k!}\frac{k^k}{(k+1)^{k+1}}=e(k+1)\frac{k^k}{(k+1)^{k+1}}=e\frac{k^k}{(k+1)^{k}}=e(\frac{k}{k+1})$ – 2017-01-10
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    @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*](https://en.wikipedia.org/wiki/Bernoulli's_inequality), or you can apply the binomial theorem to expand $(1+\frac{1}{k})^k$ and compare the resulting sum with $e = \sum_{j=0}^{\infty} \frac{1}{j!}$. – 2017-01-10
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    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. – 2017-01-10

2 Answers 2

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

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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.