# Find the power series expansion of $f(x)= \frac{e^x - 1}{x}$

Prove that the following function is analytical in 0 and find its power series centered in 0

$$f(x)= \frac{e^x - 1}{x}, f(0)=1$$

I'm trying to write $$f$$ as some kind of combination of function with known Taylor series expansion in an open interval around zero, which would prove the analicity:

$$f(x)= \frac{e^x}{x}- \frac{1}{x}=\frac{1}{x}\sum_{n=0}^{\infty} \frac{x^n}{n! } - \frac{1}{x} =\sum_{n=0}^{\infty} \frac{x^{n-1}}{n! } - \frac{1}{x}$$

but I don't know about any power series expansion of $$\frac{1}{x}$$

• Maybe look at the power series of $e^x-1$ and divide that by $x$? Feb 5 '19 at 19:32
• @hamam_Abdallah fixed, thanks Feb 5 '19 at 19:33
• @BarryCipra fixed, thank you. I shouldn't copy-paste mathjax expressions without double-checking haha Feb 5 '19 at 19:37
• Look at the $n =0$ term in your sum :) Feb 5 '19 at 19:43

$$e^x- 1 = \sum_{n=1}^\infty \frac{x^n}{n!}$$ Now divide each term by $$x$$.
$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$$ $$\to e^x-1=x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$$ $$\to \frac{e^x-1}{x}=1+\frac{x}{2!}+\frac{x^2}{3!}+...$$
$$f(x)=\frac{e^x-1}{x}=\frac{e^x}{x}-\frac{1}{x}$$ we know that $$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$$ so: $$\frac{e^x}{x}=\sum_{n=0}^\infty\frac{x^{n-1}}{n!}=\frac{1}{x}+1+\frac{x}{2}+...$$ And so: $$f(x)=1+\frac{x}{2!}+\frac{x^2}{3!}+...=\sum_{n=1}^\infty\frac{x^{n-1}}{n!}$$
You know that $$e^x= \sum_{n=0}^\infty \frac{x^n}{n!}= 1+ x+ \frac{x^1}{2!}+ \cdot\cdot\cdot+ \frac{x^n}{n!}+ \cdot\cdot\cdot$$, so $$e^x- 1= \sum_{n=1}^\infty \frac{x^n}{n!}= x+ \frac{x^1}{2!}+ \cdot\cdot\cdot+ \frac{x^n}{n!}+ \cdot\cdot\cdot.$$ thus $$\frac{e^{x}- 1}{x}= \sum_{n= 1}^\infty \frac{x^{n-1}}{n!}=\sum_{n= 0}^\infty \frac{x^n}{(n+1)!}= 1+ \frac{x}{2!}+ \frac{x^2}{3!}+ \cdot\cdot\cdot+ \frac{x^n}{n!}+ \cdot\cdot\cdot$$