Answer
$R=0.1$
$I=[-0.1,0.1]$
Work Step by Step
Given $$\sum_{n=1}^{\infty} \frac{10^nx^n}{ n^3}$$
Since $a_n= \dfrac{10^nx^n}{ n^3} $, $a_{n+1}= \dfrac{10^{n+1}x^{n+1}}{ (n+1)^3} $, then
\begin{align*}
\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|&=\lim_{n\to\infty}\left|\dfrac{10^{n+1}x^{n+1}}{ (n+1)^3} \dfrac{ n^3}{10^nx^n}\right|\\
&=\lim_{n\to\infty}\left|\dfrac{10 xn^3 }{ (n+1)^3} \right|\\
&=|10x|\lim_{n\to\infty}\left( \dfrac{ n }{ n+1 } \right)^3\\
&=|10x|
\end{align*}
By the Ratio Test, the series $\displaystyle\sum_{n=1}^{\infty} \frac{10^nx^n}{ n^3} $ converges when $ |10x|<1\ \Rightarrow -1\lt10x\lt1$. Hence the radius of convergence is $R=0.1$
At $x= 0.1$ the series $\displaystyle\sum_{n=1}^{\infty} \frac{10^nx^n}{ n^3} = \displaystyle\sum_{n=1}^{\infty} \frac{1 }{ n^3} $, is a convergent series ($p-$series $p\gt1$),
At $x= -0.1$ the series $\displaystyle\sum_{n=1}^{\infty} \frac{10^nx^n}{ n^3} = \displaystyle\sum_{n=1}^{\infty} \frac{(-1 )^n }{ n^3} $ is a convergent series.
Hence, the interval of convergence is $I=[-0.1,0.1]$.
