Question

Use the ratio test to find whether the following series converge or diverge:

24.$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{{\mathbf{3}}^{\mathbf{2}\mathbf{n}}}{{\mathbf{2}}^{\mathbf{3}\mathbf{n}}}$

Short answer

The series $\sum _{n=0}^{\infty }\frac{3^{2n}}{2^{3n}}$divergence.

Step-by-step solution

Process of ratio test

Apply ratio test in the given series by using${\mathit{\rho }}_{\mathbf{n}}\mathbf{=}\left|\frac{a_{n+1}}{a_n}\right|$and$\mathit{\rho }\mathbf{=}\underset{\mathbf{n}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}{\mathit{\rho }}_{\mathbf{n}}$, where${\mathit{a}}_{\mathbf{n}\mathbf{+}\mathbf{1}}$ is the${\left(n+1\right)}^{\mathbf{t}\mathbf{h}}$ term of the series and${\mathit{a}}_{\mathbf{n}}$ is the${\mathit{n}}^{\mathbf{t}\mathbf{h}}$ term. If $\mathit{\rho }\mathbf{<}\mathbf{1}$, then the series converges. If $\mathit{\rho }\mathbf{>}\mathbf{1}$, then the series diverges.

Apply the ratio test

The given series is $\sum _{n=0}^{\infty }\frac{3^{2n}}{2^{3n}}$.

So, $a_{n+1}=\frac{3^{2\left(n+1\right)}}{2^{3\left(n+1\right)}}$, and role="math" $a_n=\frac{3^{2n}}{2^{3n}}$.

Obtain the value of ${\rho }_n=\left|\frac{a_{n+1}}{a_n}\right|$.

$\begin{array}{rcl}{\rho }_n& =& \left|\frac{a_{n+1}}{a_n}\right|\\ & =& \left|\frac{3^{2\left(n+1\right)}}{2^{3\left(n+1\right)}}÷\frac{3^{2n}}{2^{3n}}\right|\\ & =& \frac{3^{2\left(n+1\right)}}{2^{3\left(n+1\right)}}\times \frac{2^{3n}}{3^{2n}}\\ & =& \frac{3^2}{2^3}\\ & =& \frac{9}{8}\end{array}$

Solve the limit

Now,$\rho =\underset{n\to \infty }{\lim }{\rho }_n$ is calculated as follows:

$\begin{array}{rcl}\rho & =& \underset{n\to \infty }{\lim }\frac{9}{8}\\ & =& \frac{9}{8}\end{array}$

Here, $\rho >1$, therefore the series diverges.