Question

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

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

Short answer

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

Step-by-step solution

Process of ratio test

Apply ratio test in the given series by using, ${\mathit{\rho }}_{\mathbf{n}}\mathbf{=}\mathbf{\left|}\frac{{\mathbf{a}}_{\mathbf{n}\mathbf{+}\mathbf{1}}}{{\mathbf{a}}_{\mathbf{n}}}\mathbf{\right|}$and $\mathit{\rho }\mathbf{=}\underset{\mathbf{n}\mathbf{\to }\mathbf{\infty }}{\mathbf{lim}}{\mathit{\rho }}_{\mathbf{n}}$, where ${\mathbf{a}}_{\mathbf{n}\mathbf{+}\mathbf{1}}$is the${\left(n+1\right)}^{\mathbf{th}}$ term of the series and${\mathbf{a}}_{\mathbf{n}}$ is the${\mathbf{n}}^{\mathbf{th}}$ term. If role="math" localid="1664190384146" $\mathbf{\text{ρ<1}}$, then the series converges. If role="math" $\mathbf{\text{ρ>1}}$, then the series diverges.

Apply the ratio test

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

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

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

$\begin{array}{rcl}{\rho }_n& =& \left|\frac{a_{n+1}}{a_n}\right|\\ & =& \left|\frac{3^{n+1}}{2^{2\left(n+1\right)}}÷\frac{3^n}{2^{2n}}\right|\\ & =& \frac{3^{n+1}}{2^{2\left(n+1\right)}}\times \frac{2^{2n}}{3^n}\\ & =& \frac{3}{4}\\ & & \end{array}$

Solve the limit

Now, is calculated as follows:

$$\begin{gathered} \text{ρ=}\underset{\mathrm{n}\to \infty }{\lim }{\mathrm{\rho }}_{\mathrm{n}} \\ =\underset{\mathrm{n}\to \infty }{\lim }\frac{3}{4} \\ =\frac{3}{4} \end{gathered}$$

Here, $\rho <1$, therefore the series converges.