Question

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

$\sum _{k=0}^{\infty }3{\left(\frac{-2}{5}\right)}^k$

Short answer

The sum of the series is$\frac{15}{7}$.

Step-by-step solution

Step 1. Given Information

The given series is$\sum _{k=0}^{\infty }3{\left(\frac{-2}{5}\right)}^k$.

Step 2. Determine if a geometric series 

• Find the ratio of the consecutive terms.

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

• Since the ratio of the consecutive terms is same, the given series is a geometric series with a common ratio $r=\begin{array}{rcl}& & \frac{-2}{5}\end{array}$.

Step 3. Determine the convergence 

• Determine the first term and the common ratio.

$\begin{array}{rcl}c& =& 3{\left(\frac{-2}{5}\right)}^0\\ & =& 3\\ r& =& \frac{-2}{5}\end{array}$

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

$\begin{array}{rcl}\frac{c}{1-r}& =& \frac{3}{1-\left({\displaystyle \frac{-2}{5}}\right)}\\ & =& \frac{3}{1+{\displaystyle \frac{2}{5}}}\\ & =& \frac{15}{7}\end{array}$