Question

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

18.$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{{\mathbf{2}}^{\mathbf{n}}}{{\mathbf{n}}^{\mathbf{2}}}$

Short answer

The series $\sum _{n=1}^{\infty }\frac{2^n}{n^2}$diverges.

Step-by-step solution

Process of ratio test

Apply ratio test in the given series by using,${\mathbf{\text{ρ}}}_{\mathbf{n}}\mathbf{=}\left|\frac{a_{n+1}}{a_n}\right|$and$\mathbf{\text{ρ=}}\underset{\mathbf{n}\mathbf{\to }\mathbf{\infty }}{\mathbf{lim}}{\mathbf{\rho }}_{\mathbf{n}}$, 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 $\mathbf{\text{ρ<1}}$, then the series converges. If $\mathbf{\text{ρ>1}}$, then the series diverges.

Apply the ratio test

The given series is $\sum _{n=1}^{\infty }\frac{2^n}{n^2}$.

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

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

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

Solve the limit

Now,$\text{ρ=}\underset{\mathrm{n}\to \infty }{\lim }{\text{ρ}}_{\mathrm{n}}$ is calculated as follows:

$\begin{array}{rcl}& & \text{ρ=}\underset{\mathrm{n}\to \infty }{\lim }{\text{ρ}}_{\mathrm{n}}\\ & =& \underset{\mathrm{n}\to \infty }{\lim }\frac{2{\mathrm{n}}^2}{{\left(\mathrm{n}+1\right)}^2}\\ & =& \underset{\mathrm{n}\to \infty }{\lim }\frac{2{\mathrm{n}}^2}{{\mathrm{n}}^2{\left(1+\frac{1}{\mathrm{n}}\right)}^2}\\ & =& \underset{\mathrm{n}\to \infty }{\lim }\frac{2}{{\left(1+\frac{1}{\mathrm{n}}\right)}^2}\\ & =& \frac{2}{{\left(1+0\right)}^2}\\ \mathrm{\rho }& =& 2\end{array}$

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