MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This seems like it must have been addressed somewhere already, but I cannot find it in any standard series tables.

I have the equation:

$f(z) = \left(1 + \frac{1}{z}\right)^z$.

What is the general form for the $n$th term of the series? That is, if I have

$f(z) \sim \sum_{n=0}^{\infty} \frac{c_n}{z^n}$

near $z = \infty$, what is the form of $c_n$?

share|cite|improve this question
up vote 14 down vote accepted

Markus Brede proves the following formula in the paper "On the convergence of the sequence defining Euler’s number". Let $$\left(1+\frac{1}{z}\right)^z=\sum_{n\geq 0} \frac{a_n}{z^n}$$ then we have $$a_n=e\sum_{v=0}^n \frac{S(n+v,v)}{(n+v)!}\sum_{m=0}^{n-v}\frac{(-1)^m}{m!}$$ where $S(a,b)$ are Stirling numbers of the first kind. This shows that all coefficients are rational multiples of $e$. I found the article through OEIS.

share|cite|improve this answer
Cool, who knew... – Igor Rivin Oct 6 '11 at 22:51
Thank you, that is excellent. I should have gone straight to OEIS. I can't believe that I didn't think to do so. It turns out that I also needed another related series which was easily found there. – Chris Oct 7 '11 at 9:52

$\log f(z) = z \log(1+1/z) = \sum_{k=0}^\infty \frac{(-1)^k}{k+1} z^{-k}$ as $z \to +\infty$, so $f(z)$ is the exponential of this sum. See for the numerators and for the denominators of the coefficients.

share|cite|improve this answer
There are two deleted answers which say the same thing (the authors deleted them because they don't actually answer the question as posed,unlike @Gjergji's answer...) – Igor Rivin Oct 6 '11 at 22:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.