Use the Root Test to determine the convergence or divergence of the series. ∑n=1∞(3n+2n+3)n

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

Answer is given below (on video)