Show that $1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots +\frac{x^{n}}{n!}+\cdots$ converges to $e^{x}$ for every number $x$. That is, prove that $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ e^{x}=1+\frac{x}{1!}+\frac{x^{2}}{2!}+\frac{x^{3}}{3!} +\cdots +\frac{x^{n}}{n!}+\cdots =\sum\limits_{k\,=\,0}^{\infty }\frac{ x^{k}}{k!}}}$$

for all real numbers.

Solution To prove that $1+\frac{x}{1!}+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots =e^{x}$ for every number $x$, we need to show that $\lim\limits_{n\rightarrow \infty }R_{n}(x)=0$. Since $f^{\,(n+1)}(x)=e^{x}$, we have $$R_{n}(x)=\frac{f^{(n+1)}(u) x^{n+1}}{(n+1) !}=\frac{e^{u}x^{n+1}}{(n+1)!}$$

where $u$ is between $0$ and $x.$ To show that $\lim\limits_{n\rightarrow\infty }R_{n}(x) = 0$, we consider two cases: $x>0$ and $x<0.$

Case 1: When $x>0$, then $0<u<x$, so that $1<e^{u}<e^{x}$ and, for every positive integer $n$, $$0<R_{n}(x)=\frac{e^{u}x^{n+1}}{(n+1)!}<\frac{e^{x}x^{n+1}}{(n+1)!}$$

By the Ratio Test, the series $\sum\limits_{k\,=\,\,0}^{\infty }\frac{x^{k+1}}{(k+1) !}$ converges for all $x$. It follows that $\lim\limits_{n\rightarrow \infty }$ $\frac{x^{n+1}}{(n+1) !}=0$, and, therefore, $$\lim\limits_{n\rightarrow \infty }\frac{e^{x}x^{n+1}}{(n+1)!} =e^{x}\lim\limits_{n\rightarrow \infty }\frac{x^{n+1}}{(n+1)!}=0$$

By the Squeeze Theorem, $\lim\limits_{n\rightarrow \infty }R_{n}(x)=0$.

Case 2: When $x<0,$ then $x<u<0$ and $e^{x}<e^{u}<1$, so that $$0\leq \left\vert R_{n}(x)\right\vert =\frac{e^{u}\cdot |x|^{n+1}}{(n+1)!}< \frac{|x|^{n+1}}{(n+1)!}$$

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Since $\lim\limits_{n\rightarrow \infty }\frac{\vert x\vert^{n+1}}{(n+1) !}=0,$ by the Squeeze Theorem, $\lim\limits_{n\rightarrow \infty }R_{n}(x)=0$. So for all $x$, $\lim\limits_{n\rightarrow \infty }R_{n}(x) =0.$ As a result, $$e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots +\frac{x^{n}}{n!}+\cdots =\sum_{k\,=\,0}^{\infty }\frac{x^{k}}{k!}$$ for all numbers $x.$