I have the following series: $$\sum_{k=1}^{\infty}\frac{k!\exp(k)}{(k+1)^k}.$$

I want to prove that it diverges.

I have already tried Alembert's quotient test and square root test and they are inconclusive. I have not found any suitable series for a comparison test or $k^a$ test.

I think $$\lim_{k\to\infty} \frac{k!\exp(k)}{(k+1)^k}= +\infty,$$ but I cannot prove it.

Any help or suggestions?


3 Answers 3


$\def\e{\mathrm{e}}$By Stirling's formula,$$ k! \sim \sqrt{2πk} \left(\frac{k}{\e}\right)^k, \quad (k \to \infty) $$ thus$$ \frac{k!\,\e^k}{(k + 1)^k} \sim \sqrt{2πk} \left(\frac{k}{\e}\right)^k \cdot \frac{\e^k}{(k + 1)^k} = \sqrt{2πk} \left(1 - \frac{1}{k + 1}\right)^k \sim \frac{\sqrt{2πk}}{\e}, \quad (k \to \infty) $$ which implies that $\sum\limits_{k = 1}^\infty \dfrac{k!\,\e^k}{(k + 1)^k}$ diverges.


Without Stirling:

$\frac{a_k}{a_{k-1}}=\frac{e}{\left(1+\frac1k\right)^k}>1$ and so $a_k$ doesn't go to $0.$

  • $\begingroup$ Could you be a bit more explicit, beacause i already tried that test and it dodnt work quite like that...? $\endgroup$
    – Desperados
    Mar 4, 2018 at 1:49
  • $\begingroup$ But isn't that $e \over e$ so the ratio test is inconclusive? $\endgroup$
    – Andrew Li
    Mar 4, 2018 at 2:04
  • $\begingroup$ To the proposer: Check your calculation. To the solver: Excellent elementary 1-liner. $\endgroup$ Mar 4, 2018 at 2:05
  • $\begingroup$ @Ilias Koutroumpas, $(1+\frac1k)^k$ tends to $e$ increasing, so for any $k$ we have $(1+\frac1k)^k<e$. This implies $a_{k+1}>a_k$ and so $a_k$ doesn't go to 0. $\endgroup$
    – Minz
    Mar 4, 2018 at 3:51

Just added for you curiosity.

As Alex Francisco answered, using Stirling approximation shows that the summation diverges.

We could even approximate the partial sum $$S_p=\sum_{k=1}^{p}\frac{k!\, e^k}{(k+1)^k}=\sum_{k=1}^{p}a_k$$ Using again Stirling approximation, $$\log(a_k)=\log \left(\frac{\sqrt{2 \pi }}{e}\right)+\frac{1}{2} \log \left({k}\right)+\frac{7}{12 k}+O\left(\frac{1}{k^2}\right)$$ Continuing with Taylor $$a_k=e^{\log(a_k)}=\frac{\sqrt{2 \pi } }{e}\sqrt{k}+\frac{7 \sqrt{\frac{\pi }{2}} }{6 e}\sqrt{\frac{1}{k}}+O\left(\frac{1}{k}\right)$$ making $$S_p=\frac{\sqrt{\frac{\pi }{2}}}{6 e}\left(12 H_p^{\left(-\frac{1}{2}\right)}-7 \zeta \left(\frac{1}{2},p+1\right)+7 \zeta \left(\frac{1}{2}\right)\right)$$ where appear generalized harmonic numbers and the Hurwitz zeta function.

Using asymptotics, $$S_p=\frac{2 \sqrt{2 \pi }}{3 e} p^{3/2}+\frac{5 \sqrt{2 \pi } }{3 e}p^{1/2}+\frac{\sqrt{\frac{\pi }{2}} \left(12 \zeta \left(-\frac{1}{2}\right)+7 \zeta \left(\frac{1}{2}\right)\right)}{6 e}+O\left(\frac{1}{p^{1/2}}\right)$$ in which the constant term is almost $-1$ ($\approx -0.977244$).

So, let us use $$S_p \approx \frac{2 \sqrt{2 \pi }}{3 e} p^{3/2}+\frac{5 \sqrt{2 \pi } }{3 e}p^{1/2}-1$$ and below are given some results

$$\left( \begin{array}{ccc} p & \text{approximation} & \text{exact} \\ 10 & 23.3004 & 23.1897 \\ 100 & 629.127 & 628.889 \\ 1000 & 19488.0 & 19487.7 \\ 10000 & 614911. & 614910. \end{array} \right)$$


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