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

Find the radius of convergence and interval of convergence of the series.
$\sum _{n=1}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{n}{10}^{n}{x}^{n}}{{n}^{4}}$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

cyhuddwyr9

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