1
$\begingroup$

I'm trying to prove that the series below converges to 1 and I noticed it looked strikingly similar to a probability distribution I once saw. My question is twofold:

  1. Can anyone identify the distribution? I can't seem to, for the life of me, remember. I'm 90% sure this is a probability distribution (or some form of one) but I may be wrong.
  2. Does anyone have any hints for proving this convergence? I don't necessarily want an answer, just some pointers. I don't think this requires any complex mathematics beyond an early graduate course in analysis.

$$ \sum_{j=1}^{\infty}{\frac{e^{-j}j^{j-1}}{j!}}= 1 $$

Thank you!

$\endgroup$
2
  • 2
    $\begingroup$ The identity follows by setting $x=1$ in the well-known formula $$\sum_{j=1}^\infty \frac{(xe^{-x})^j j^{j-1}}{j!} = x,$$ which is equivalent to the Taylor series for the Lambert W function that Robert Israel mentioned. A simple proof is by expanding the left side in powers of $x$ and using the fact that the $n$th difference of a polynomial of degree less than $n$ is 0. Another simple proof was given by Noam Elkies, math.harvard.edu/~elkies/Misc/abel.pdf. $\endgroup$
    – Ira Gessel
    Commented Dec 25, 2016 at 17:31
  • $\begingroup$ @IraGessel That's a really nice proof. Thank you. If you'll make your comment into an answer I will gladly up vote it. $\endgroup$
    – gowrath
    Commented Dec 25, 2016 at 22:03

1 Answer 1

2
$\begingroup$

1) Presumably you're talking about the Poisson distribution, but I don't think that's useful.

2) Possibly more useful is that this is related to the Taylor series for the Lambert W function.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .