Does the infinite geometric series converge or diverge? 1 + 4/3 + 16/9 + 64/27 + Choices: (A)converge (B)diverge

Find Common Ratio: To determine if the infinite geometric series converges or diverges, we need to find the common ratio (r) of the series. The common ratio is found by dividing any term in the series by the term preceding it. Let's take the second term (4/3) and divide it by the first term (1). r = (4/3) / 1 = 4/3Calculate Sum Formula: Now that we have the common ratio r = 4/3, we can use the formula for the sum of an infinite geometric series, which is S = a / (1 - r), where S is the sum, a is the first term, and r is the common ratio. This formula only applies if the absolute value of r is less than 1. Since the common ratio r = 4/3 is greater than 1, the absolute value of r is not less than 1.Determine Convergence: Because the absolute value of the common ratio is greater than 1, the series does not meet the criteria for convergence. Therefore, the infinite geometric series diverges.

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Does the infinite geometric series converge or diverge? $\newline$ $1 + \frac{4}{3} + \frac{16}{9} + \frac{64}{27} + \ldots$ $\newline$ Choices: $\newline$ (A) converge $\newline$ (B) diverge

View step-by-step help

Home

Math Problems

Precalculus

Convergent and divergent geometric series

Full solution

Q. Does the infinite geometric series converge or diverge?

\newline

1 + \frac{4}{3} + \frac{16}{9} + \frac{64}{27} + \ldots

\newline

Choices:

\newline

(A) converge

\newline

(B) diverge

Find Common Ratio: To determine if the infinite geometric series converges or diverges, we need to find the common ratio $r$ of the series. $\newline$ The common ratio is found by dividing any term in the series by the term preceding it. $\newline$ Let's take the second term $\frac{4}{3}$ and divide it by the first term $1$ . $\newline$ $r = \frac{\frac{4}{3}}{1} = \frac{4}{3}$
Calculate Sum Formula: Now that we have the common ratio $r = \frac{4}{3}$ , we can use the formula for the sum of an infinite geometric series, which is $S = \frac{a}{1 - r}$ , where $S$ is the sum, $a$ is the first term, and $r$ is the common ratio. $\newline$ This formula only applies if the absolute value of $r$ is less than $1$ . $\newline$ Since the common ratio $r = \frac{4}{3}$ is greater than $1$ , the absolute value of $r$ is not less than $1$ .
Determine Convergence: Because the absolute value of the common ratio is greater than $1$ , the series does not meet the criteria for convergence. $\newline$ Therefore, the infinite geometric series diverges.