Find all numbers \(x\) for which each power series in \(x\) converges.
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Then \[ \lim\limits_{n\rightarrow \infty }\left\vert \frac{a_{n+1}}{a_{n}}\right\vert =\lim\limits_{n\rightarrow \infty }\left\vert \frac{\dfrac{x^{n+1}}{(n+1) !}}{\dfrac{x^{n}}{n!}}\right\vert =\lim\limits_{n\rightarrow \infty } \frac{|x|^{n+1}\,n!}{(n+1)!\,|x|^{n}}=\vert x\vert \lim\limits_{n\rightarrow \infty }\frac{1}{n+1}=0 \]
Since the limit is less than \(1\) for every number \(x\), it follows from the Ratio Test that the power series \(\sum\limits_{k\,=\,0}^{\infty }\dfrac{x^{k}}{k!}\) is absolutely convergent for all real numbers.
Since \(\sum\limits_{k\,=\,0}^{\infty} \dfrac{{ x}^{k}}{{k!}}\) converges absolutely for every number \(x\), the limit of the \(n\)th term equals \({0}\). That is, \(\lim\limits_{n\,\rightarrow \,\infty }\dfrac{{x}^{n}}{{n!}}={ 0}\) for every number \(x\).
(b) For \(\sum\limits_{k\,=\,0}^{\infty }\dfrac{kx^{k}}{4^{k}}\), we use the Ratio Test with \(a_{n}=\dfrac{nx^{n}}{4^{n}}\) and \(a_{n+1}=\dfrac{ (n+1)x^{n+1}}{4^{n+1}}\). Then \[ \begin{eqnarray*} \lim\limits_{n\,\rightarrow \,\infty }\left\vert \frac{a_{n+1}}{a_{n}} \right\vert &=& \lim\limits_{n\,\rightarrow \,\infty }\dfrac{\dfrac{ (n+1)\,\vert x\vert ^{n+1}}{4^{n+1}}}{\dfrac{n\,\,\vert x\vert ^{n}}{4^{n}}}=\lim\limits_{n\,\rightarrow \,\infty }\frac{ (n+1)\,|x|^{n+1}(4^{n})}{4^{n+1}\cdot n\,|x|^{n}} \\ &=& \vert x\vert \,\lim\limits_{n\,\rightarrow \,\infty }\frac{n+1}{4n}=\frac{|x|}{4} \end{eqnarray*} \]
By the Ratio Test, the series converges absolutely if \(\dfrac{|x|}{4} < 1\), or equivalently if \(\vert x\vert < 4\). It diverges if \(\dfrac{|x|}{4} \gt 1\) or equivalently if \(\vert x\vert >4\). The Ratio Test gives no information when \(\dfrac{|x|}{4}=1\), that is, when \(x=-4\) or \(x=4\). However, we can check these values directly by replacing \(x\) by \(4\) and \(-4\).
For \(x=4\), the series becomes \[ \sum_{k\,=\,0}^{\infty }\frac{k4^{k}}{4^{k}}=\sum_{k=1}^{\infty }k=1+2+\cdots \]
which diverges. For \(x=-4\), the series becomes \[ \sum_{k=0}^{\infty }\frac{k(-4)^{k}}{4^{k}}=\sum_{k=0}^{\infty }\frac{ (-1)^{k}k(4^{k})}{4^{k}}=\sum_{k=0}^{\infty }(-1)^{k}k=-1+2-3+\cdots \]
which also diverges. (Look at the sequence of partial sums.)
The series \(\sum\limits_{k\,=\,0}^{\infty }\dfrac{kx^{k}}{4^{k}}\) converges absolutely for \( -4< x< 4\) and diverges for \(\vert x\vert \geq 4\).
(c) For \(\sum\limits_{k\,=\,0}^{\infty }k!x^{k}\), we use the Ratio Test with \(a_{n}=n!\,x^{n}\) and \(a_{n+1}=(n+1)!\,x^{n+1}\). Then \[ \lim\limits_{n\,\rightarrow \,\infty }\left\vert \frac{a_{n+1}}{a_{n}} \right\vert =\lim\limits_{n\,\rightarrow \,\infty }\frac{(n+1)!\,\vert x\vert ^{n+1}}{n\dot{!}\,\vert x\vert ^{n}}=\vert x\vert\,\lim\limits_{n\,\rightarrow \,\infty } (n+1) =\left\{ \begin{array}{ll} 0 & \hbox{if} \, x=0 \\[4pt] \infty & \hbox{if} \, x\neq 0 \end{array} \right. \]
We conclude that the power series \(\sum\limits_{k\,=\,0}^{\infty }k!x^{k}\) converges only when \(x=0\). For any other number \(x\), the power series diverges.