Question

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

$\sum _{k=0}^{\infty }\frac{5^{k+3}}{2^{3k+1}}$

Short answer

The sum of the series is$\frac{500}{3}$.

Step-by-step solution

Step 1. Given Information

The given series is$\sum _{k=0}^{\infty }\frac{5^{k+3}}{2^{3k+1}}$.

Step 2. Determine if a geometric series   

• Find the ratio of the consecutive terms.

$\begin{array}{rcl}\frac{{\displaystyle \frac{5^{k+3+1}}{2^{3(k+1)+1}}}}{\frac{5^{k+3}}{2^{3\left(k\right)+1}}}& =& \frac{5^{k+3+1-k-3}}{2^{3(k+1)+1-3k-1}}\\ & =& \frac{5}{2^3}\\ & =& \frac{5}{8}\end{array}$

• Since the ratio of the consecutive terms is same, the given series is a geometric series with a common ratio $r=\begin{array}{rcl}& & \frac{5}{8}\end{array}$.

Step 3. Determine the convergence   

• Determine the first term and the common ratio.

$\begin{array}{rcl}c& =& \frac{5^{0+3}}{2^{3\left(0\right)+1}}\\ & =& \frac{5^3}{2}\\ & =& \frac{125}{2}\\ r& =& \frac{5}{8}\end{array}$

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

$\begin{array}{rcl}\frac{c}{1-r}& =& \frac{{\displaystyle \frac{125}{2}}}{1-{\displaystyle \frac{5}{8}}}\\ & =& \frac{{\displaystyle \frac{125}{2}}}{{\displaystyle \frac{3}{8}}}\\ & =& \frac{500}{3}\end{array}$