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.
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 \(x<0.\)
The Squeeze Theorem is discussed in Section 1.4, pp. 106-107.
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 \(x.\)