Find the radius of convergence and interval of convergence of the series. sum_(n=1)^oo((-1)^n10^nx^n)/n^4

Here ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n{10}^n{x}^n}{n^4}=\sum _{n=1}^{\mathrm{\infty }}{a}_n,where{a}_n=\frac{(-1{)}^n{10}^n{x}^n}{n^4}}$
Now ${\displaystyle \left|\frac{a_n+1}{a_n}\right|=\left|\frac{{10}^{n+1}{x}^{n+1}{n}^2}{{10}^n{x}^n{(n+1)}^2}\right|}$
${\displaystyle =\left|10x{\left(\frac{n}{n+1}\right)}^2\right|}$
This shows that
${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_n+1}{a_n}\right|=\left|10x\right|\underset{n\to \mathrm{\infty }}{lim}{\left(\frac{n}{n+1}\right)}^2=10\left|x\right|}$
Therefore the given series converges if
${\displaystyle 10\left|x\right|<1\Rightarrow \left|x\right|<\frac{1}{10}}$
and diverges if ${\displaystyle \left|x\right|>\frac{1}{10}}$. Therefore the radious of converges is ${\displaystyle R=\frac{1}{10}}$.When ${\displaystyle x=\frac{1}{10}}$ the the series beomes ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^4}}$ which converges by the Leibniz's test. Again when ${\displaystyle x=-\frac{1}{10}}$ then the series becomes ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^4}}$ which converges by the p series test. Thereore the series converges converges in the interval ${\displaystyle \left|x\right|\le \frac{1}{10}}$