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Is there a formula for the coefficients in the expansion of $\frac1{e}(1+1/x)^x$ as $x \to \infty$?

This is inspired by Evaluate: $ \lim_{x\to 0}\frac{(1+x)^{1/x}-e+\dfrac{1}{2}ex}{x^2}$

According to Wolfy, the first few terms are

$1 - 1/(2 x) + (11 )/(24 x^2) - (7 )/(16 x^3) + O((1/x)^4) $.

Is there a not too complex formula, possibly involving some nested summations, for the expansion

$\frac1{e}(1+1/x)^x =\sum_{n=0}^{\infty} a_n/x^n $?

I would not be surprised if this has been asked before, but it did not appear in the list of similar questions.

marty cohen
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1 Answers1

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A formula for such coefficients $(a_n)_{n\geq 0}$ is given in terms of Stirling numbers of the first kind: $$a_n=\sum_{k=0}^n\frac{s(n+k,k)}{(n+k)!}\sum_{j=0}^{n-k}\frac{(-1)^j}{j!}$$ See OEIS-A055505 for references. The first terms are $$1, -\frac{1}{2}, \frac{11}{24}, -\frac{7}{16}, \frac{2447}{5760}, -\frac{959}{2304}, \frac{238043}{580608}, -\frac{67223}{165888}, \frac{559440199}{1393459200},\dots$$

Robert Z
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