Exercise 3.9

Question

Exercise 9:
    Find the radius of convergence of each of the following power series: $$\hbox{(a) }\sum n^3z^n,\quad\quad\hbox{(b) }\sum\frac{2^n}{n!}z^n,\quad\quad\hbox{(c) }
    \sum\frac{2^n}{n^2}z^n,\quad\quad\hbox{(d) }\sum \frac{n^3}{3^n}z^n.$$

ghostofgarborg · Accepted Answer

(a) $$ \frac 1 R = \lim_{n \to \infty} \frac{(n+1)^3}{n^3} = \lim_{n \to \infty} ( 1 + \frac 3 n + \frac 3 {n^2} + \frac 1 {n^3} ) = 1 $$
    So $R = 1$.

(b) 
    $$ \frac 1 R = \lim_{n \to \infty} \frac{\frac{2^{n+1}}{(n+1)!}}{\frac{2^n}{n!}} = \lim_{n \to \infty} \frac 2 {n+1} = 0 $$
    So $R = \infty$.

(c)
    $$ \frac 1 R = \lim_{n \to \infty} \sqrt[n]{\frac{2^n}{n^2}} = \lim_{n \to \infty} \frac{2}{(\sqrt[n]{n})^2} = 2 $$ 
    So $R = \frac 1 2$.

(c)
    $$ \frac 1 R = \lim_{n \to \infty} \sqrt[n]{\frac{n^3}{3^n}} = \lim_{n \to \infty} \frac{(\sqrt[n]{n})^3}{3} = \frac 1 3 $$ 
    So $R = 3$.

Round · Answer

Another way to solve (b):
For each $m$, if $n\geq m^2+m$, then
$$
\begin{aligned}
n!&=m!\cdot(m+1)\cdot \cdots \cdot m^2 \cdot (m^2+1) \cdot \cdots \cdot (m^2+m) \cdot \cdots \cdot n\
 &>1^m \cdot m^{m^2-m} \cdot (m^2)^m\cdot m^{n-m^2-m}\
&=m ^n,
\end{aligned}
$$
hence $\sqrt[n]{n!}>m$.
This implies 
$$\lim_{nightarrow \infty}\sqrt[n]{\frac{2^n}{n!}}=\lim_{nightarrow \infty}\frac{2}{\sqrt[n]{n!}}=0.$$
So $R=\infty.$

Exercise 3.9

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Exercise 3.9

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