Marvin Mccormick

2020-11-27

Find the region of convergence and absolute convergence of the series.
$\sum _{n=1}^{\mathrm{\infty }}\frac{{\mathrm{ln}}^{n}x}{n}$

Brighton

Given the series, $\sum _{n=1}^{\mathrm{\infty }}\frac{{\mathrm{ln}}^{n}x}{n}$
Here, $|{a}_{n}|=|\frac{\left(\mathrm{ln}x{\right)}^{n}}{n}|$
To find the region of convergence and absolute convergence of the given series.
By ratio test, if $\underset{n\to \mathrm{\infty }}{lim}\frac{{a}_{n+1}}{{a}_{n}}<1$, the series converges.
Therefore, the absolute convergence of the series implies,
$\underset{n\to \mathrm{\infty }}{lim}|\frac{{a}_{n+1}}{{a}_{n}}|=\underset{n\to \mathrm{\infty }}{lim}|\frac{\left(\mathrm{ln}x{\right)}^{n+1}}{n+1}×\frac{n}{\left(\mathrm{ln}x{\right)}^{n}}|$
$=\underset{n\to \mathrm{\infty }}{lim}|\frac{n\mathrm{ln}x}{n+1}|$
$=\underset{n\to \mathrm{\infty }}{lim}|\frac{\mathrm{ln}x}{1+\frac{1}{n}}|$
$=|\mathrm{ln}x|$
So, the series absolutely converges if, $|\mathrm{ln}x|<1$
Therefore, the region of convergence is, $|\mathrm{ln}x|<1$

Jeffrey Jordon