Question

Geometric series: For each of the series that follow, find the sum or explain why the series diverges.

$\sum _{k=4}^{\infty }{\left(\frac{1}{3}\right)}^{k+2}$

Short answer

The sum of the series is$\begin{array}{rcl}& & \frac{1}{2}{\left(\frac{1}{3}\right)}^5\end{array}$.

Step-by-step solution

Step 1. Given Information

The given series is$\sum _{k=4}^{\infty }{\left(\frac{1}{3}\right)}^{k+2}$.

Step 2. Determine if a geometric series

• Find the ratio of the consecutive terms.

$\begin{array}{rcl}\frac{{\left({\displaystyle \frac{1}{3}}\right)}^{k+2+1}}{{\left(\frac{1}{3}\right)}^{k+2}}& =& {\left(\frac{1}{3}\right)}^{\left(k+2+1\right)-\left(k+2\right)}\\ & =& {\left(\frac{1}{3}\right)}^1\\ & =& \frac{1}{3}\end{array}$

• Since the ratio of the consecutive terms is same, the given series is a geometric series with a common ratiorole="math" localid="1649256775632" $r=\frac{1}{3}$.

Step 3. Determine the convergence

• Determine the first term and the common ratio.

$\begin{array}{rcl}c& =& {\left(\frac{1}{3}\right)}^{4+2}\\ & =& {\left(\frac{1}{3}\right)}^6\\ r& =& \frac{1}{3}\end{array}$

• If $\left|r\right|<1$, the series converges to $\frac{c}{1-r}$.

$\begin{array}{rcl}\frac{c}{1-r}& =& \frac{{\left({\displaystyle \frac{1}{3}}\right)}^6}{1-\left({\displaystyle \frac{1}{3}}\right)}\\ & =& \frac{{\left({\displaystyle \frac{1}{3}}\right)}^6}{\left({\displaystyle \frac{2}{3}}\right)}\\ & =& \frac{1}{2}{\left(\frac{1}{3}\right)}^5\end{array}$