Find all numbers $x$ for which each power series in $x$ converges.

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Solution (a) For the series $\sum\limits_{k\,=\,0}^{\infty }\dfrac{x^{k}}{k!}$, we use the Ratio Test with $$a_{n}=\frac{x^{n}}{n!}\qquad\hbox{and}\qquad a_{n+1}=\frac{x^{n+1}}{(n+1)!}$$

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.

(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.