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I am trying to determine the power series of $f(x)=\frac{e^{x}-1}{x}$ (suppose $f(0)=0$). I know that $e^x=\sum_{k=0}^\infty\frac{x^k}{k!}$ and $\frac{1}{x}=\sum_{k=0}^\infty(-1)^k(x-1)^k$ for $(1-x)<1$, but I can't just combine them can I? I tried combining them and then checked the definition of the n-th derivative at $x=0$ of a power series to check if it matched the expected derivative, but it didn't.

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  • $\begingroup$ There is no power seris but there is a Laurent series Just divide each term in the series for $e^{x}$ by $x$. $\endgroup$ – Kavi Rama Murthy Feb 11 at 9:48
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    $\begingroup$ $$\frac{e^x-1}{x}=\sum _{n=0}^{\infty } \frac{x^n}{(n+1)!}$$ $\endgroup$ – Raffaele Feb 11 at 10:04
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Recall that the Malaurin series for $e^x$ is $$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots$$ Hence, $$\begin{align} &e^x-1=\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots\\ \implies &\frac{e^x-1}{x}=\frac{1}{1!}+\frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}+\cdots\\ \end{align}$$ So $$\frac{e^x-1}{x}=\sum_{r=0}^\infty\frac{x^r}{(r+1)!}$$

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Though direct construction from the development of the exponential is the easy way, we can go for the explicit evaluation of the derivatives. To make the computation manageable, we start from

$$xy=e^x-1,$$

then

$$y+xy'=e^x,\\2y'+xy''=e^x,\\3y''+xy'''=e^x,\\\cdots$$ and setting $x=0$,

$$y_0=1,\\2y'_0=1,\\3y''_0=1,\\\cdots$$

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    $\begingroup$ +1 Very nice method, thank you! $\endgroup$ – A-Level Student Feb 11 at 11:02

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