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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 $c_n$?

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2 Answers

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.

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Cool, who knew... –  Igor Rivin Oct 6 '11 at 22:51
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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
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$\log f(z) = z \log(1+1/z) = \sum_{k=0}^\infty \frac{(-1)^k}{k+1} z^{-k}$ as $z \to +\infty$, so $f(z)$ is the exponential of this sum. See http://oeis.org/A055505 for the numerators and http://oeis.org/A055535 for the denominators of the coefficients.

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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
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