babeeb0oL

2021-02-21

Use the Root Test to determine the convergence or divergence of the series.
$\sum _{n=1}^{\mathrm{\infty }}\left(\frac{3n+2}{n+3}{\right)}^{n}$

### Answer & Explanation

doplovif

The given series is $\sum _{n=1}^{\mathrm{\infty }}\left(\frac{3n+2}{n+3}{\right)}^{n}$
Ratio test:
The series $\sum _{n=1}^{\mathrm{\infty }}{a}_{n}$ converges if $\underset{n\to \mathrm{\infty }}{lim}\sqrt[n]{|{a}_{n}|}<1$
The series $\sum _{n=1}^{\mathrm{\infty }}{a}_{n}$ diverges if $\underset{n\to \mathrm{\infty }}{lim}\sqrt[n]{|{a}_{n}|}>1$
In the given series $\sum _{n=1}^{\mathrm{\infty }}\left(\frac{3n+2}{n+3}{\right)}^{n},{a}_{n}=\left(\frac{3n+2}{n+3}{\right)}^{n}$
Find the limit $\underset{n\to \mathrm{\infty }}{lim}\sqrt[n]{|{a}_{n}|}$
$\underset{n\to \mathrm{\infty }}{lim}\sqrt[n]{|{a}_{n}|}=\underset{n\to \mathrm{\infty }}{lim}|\left(\frac{3n+2}{n+3}{\right)}^{n}{|}^{\frac{1}{n}}$
$=\underset{n\to \mathrm{\infty }}{lim}\left(\frac{3n+2}{n+3}\right)$
$=\underset{n\to \mathrm{\infty }}{lim}\left(\frac{3+\frac{2}{n}}{1+\frac{3}{n}}\right)$
$=\frac{3+\frac{2}{\mathrm{\infty }}}{1+\frac{3}{\mathrm{\infty }}}$
$=\frac{3+0}{1+0}$
$=3$
$>1$
By the Ratio test, the series $\sum _{n=1}^{\mathrm{\infty }}\left(\frac{3n+2}{n+3}{\right)}^{n}$ diverges.

Jeffrey Jordon