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Q 58.

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Found in: Page 825

### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

# In Problems 51–66, determine whether each infinite geometric series converges or diverges. If it converges, find its sum.$9+12+16+\frac{64}{3}+....$

The given infinite geometric series $9+12+16+\frac{64}{3}+....$ is diverges.

See the step by step solution

## Step 1. Write the given information.

The given geometric series is:

$9+12+16+\frac{64}{3}+....$

## Step 2: Find the common ratio.

${a}_{1}=9,{a}_{2}=12,{a}_{3}=16,{a}_{4}=\frac{64}{3}$

The common ratio is the ratio of successive terms:

$\frac{12}{9}=\frac{4}{3},\frac{16}{12}=\frac{4}{3}\phantom{\rule{0ex}{0ex}}r=\frac{4}{3}$

## Step 3. Determine whether the infinite geometric series is converges or diverges.

If $\left|r\right|<1$ then the infinite geometric series $\sum _{k=1}^{\infty }{a}_{1}{r}^{k-1}$ converges.

As we can see that $r=\frac{4}{3}\mathrm{and}\frac{4}{3}>1$, therefore the given series is not converges.

The given infinite geometric series is diverges.