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If it were $e^z$ being expanded at $z=1$, it would be relatively easy. We would take advantage of the fact that $$e^z =\sum_{-\infty}^\infty a_n(z-1)^n$$ where $a_n$ is given by the closed-loop integral $$a_n = \frac{1}{2\pi i} \int_\gamma \frac{e^z}{(z-1)^{n+1}}dz$$ If $n+1\leq0$, then the function inside of the integral is completely defined, so $a_n=0$. So we only worry about when $n \geq 0$. For this, we take advantage of Cauchy's integral formula to get $$a_n = \frac{e}{n!}$$ And hence $$e^z = \sum_{n=0}^\infty e(z-1)^n$$ at $z=1$


I wanted to apply the same system to evaluating $e^{1/z}$ at $z=1$. But unfortunately, $\frac{d}{dz}e^{1/z} \neq e^{1/z}$, so the Cauchy integral formula will not be of use. Is there any trick for evaluating this one?

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    $\begingroup$ There are two Laurent series for $e^{1/z}$ centred at $z=1$: one of them converges for $|z-1|<1$, the other converges for $|z-1|>1$. Which one are you interested in? $\endgroup$ Nov 12, 2022 at 19:01

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Let $0<r<1$ and $\gamma$ be the counterclockwise contour $|z-1|=r$. Then we have for $n\ge 1$ and $|z-1|<1$

$$\begin{align} a_n&=\frac1{2\pi i}\oint_{\gamma}\frac{e^{1/z}}{(z-1)^{n+1}}\,dz\\\\ &\overbrace{=}^{z=1/w}\frac{(-1)^n}{2\pi i}\oint_{\gamma'}\frac{w^{n-1}e^w}{(w-1)^{n+1}}\,dw\\\\ &=\frac{(-1)^n}{n!} \lim_{w\to 1}\frac{d^n(w^{n-1}e^w)}{dw^n}\\\\ &=\frac{(-1)^n}{n!} \lim_{w\to 1}\sum_{k=0}^n \binom{n}{k}\frac{d^{n-k}e^w}{dw^{n-k}}\frac{d^kw^{n-1}}{dw^k}\\\\ &=\frac{(-1)^n e}{n!}\sum_{k=0}^n \binom{n}{k}\frac{(n-1)!}{(n-1-k)!}\\\\ &=(-1)^n (n-1)! e \sum_{k=0}^{n-1}\frac{1}{k!(n-k)!(n-1-k)!} \end{align}$$

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Remark. A few terms of the two series:
For $|z-1|<1$, $$ e^{1/z} = {\rm e}-{\rm e} \left( z-1 \right) +{\frac {3\,{\rm e}}{2}} \left( z- 1 \right) ^{2}-{\frac {13\,{\rm e}}{6}} \left( z-1 \right) ^{3}+{ \frac {73\,{\rm e}}{24}} \left( z-1 \right) ^{4}-{\frac {167\,{\rm e} }{40}} \left( z-1 \right) ^{5}+O \left( \left( z-1 \right) ^{6} \right) $$ For $|z-1|>1$, $$ e^{1/z} = 1+ \frac{1}{\left( z-1 \right)}-{\frac {1}{2\, \left( z-1 \right) ^{2}}}+{ \frac {1}{6\, \left( z-1 \right) ^{3}}}+{\frac {1}{24\, \left( z-1 \right) ^{4}}}-{\frac {19}{120\, \left( z-1 \right) ^{5}}}+O \left( \left( z-1 \right) ^{-6} \right) $$

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  • $\begingroup$ Hello my friend! How are you? Interestingly, there is in fact a closed-form result. $\endgroup$
    – Mark Viola
    Nov 12, 2022 at 19:36
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Concerning the first series given by @GEdgar, we have $$e^{\frac{1}{z}}=e\sum_{n=0}^\infty a_n\,(z-1)^n$$ where the $a_n$ are given by $$n a_n+(2 n-1) a_{n-1}+(n-2) a_{n-2}=0$$ with $a_0=1$ and $a_1=-1$

In absolute value, the numerators form sequence $A067764$ in $OEIS$; denominators form sequence $A067653$.

In terms of special functions $$a_n=\frac {U(-n,0,-1)}{n!}$$ $U(a,b,z)$ being known as the Kummer's function of the second kind, Tricomi function, or Gordon function. This is a concise form of @Mark Viola's formula.

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