I was computing some calculations, when I got stuck about a possible closed form for this series:

$$S = \sum_{k = 2}^{N}\ \frac{k!}{k^k - k!}$$

I proved by hands that it's absolutely convergent by using the ratio test, but now I have to find something with which to approximate that series, or a closed form if it does exist.

Computing the first $50$ term of the series, the result is around

$$S \approx 1.454864034404(...)$$

If I try to take the limit for large $k$ (namely for large $N$), I can use Stirling approximation

$$k! \approx \sqrt{2\pi k} \frac{k^k}{e^k}$$

I will obtain:

$$S(\text{large}\ k) = \frac{\sqrt{2\pi k}}{e^k - \sqrt{2\pi k}}$$

Numerical evaluation

"My" alternative form seems good because even for $k = 5$ I got these:

$$ \begin{cases} \frac{5!}{5^5 - 5!} & \approx 0.03993 \\\\ \frac{\sqrt{2\pi 5}}{e^5 - \sqrt{2\pi 5}} & \approx 0.03924 \end{cases} $$

Anyway: is there a way to obtain a closed form for a general $N$? I also tried with mathematica but it doesn't work.

  • $\begingroup$ Your last sentence makes me think that there's no close form. Mathematica is usually pretty smart on this kind of things. But I can't tell for 100% of course. $\endgroup$ – Kaster Feb 7 '16 at 0:04
  • $\begingroup$ isc.carma.newcastle.edu.au/advancedCalc also finds nothing. $\endgroup$ – barrycarter Feb 7 '16 at 1:07
  • $\begingroup$ Sophomore's dream might help... $\endgroup$ – Alex Feb 7 '16 at 1:53
  • $\begingroup$ Even the simpler series $\displaystyle\sum_{k=1}^\infty\frac{k!}{k^k}$ is not known to possess a closed form. See OEIS A$94082$. $\endgroup$ – Lucian Feb 7 '16 at 2:02
  • $\begingroup$ @Lucian Indeed! It's so strange! Let's give Mathematica some lesson :D :D $\endgroup$ – Von Neumann Feb 7 '16 at 11:57

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