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

Identify Terms: To determine if the infinite geometric series converges or diverges, we need to identify the first term (a) and the common ratio (r) of the series. The first term a = 1. The common ratio r is the factor by which each term is multiplied to get the next term. In this series, each term is 1/4 of the previous term, so r = 1/4.Determine Common Ratio: An infinite geometric series converges if the absolute value of the common ratio is less than 1 (|r| < 1). It diverges if the absolute value of the common ratio is greater than or equal to 1 (|r| ≥ 1). In this case, |r| = |1/4| = 0.25, which is less than 1.Check Convergence: Since the absolute value of the common ratio is less than 1, the series converges. Therefore, the correct choice is (A) converge.

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? $\newline$ $1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \ldots$ $\newline$ 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{1}{4} + \frac{1}{16} + \frac{1}{64} + \ldots

\newline

Choices:

\newline

(A) converge

\newline

(B) diverge

Identify Terms: To determine if the infinite geometric series converges or diverges, we need to identify the first term $a$ and the common ratio $r$ of the series. $\newline$ The first term $a = 1$ . $\newline$ The common ratio $r$ is the factor by which each term is multiplied to get the next term. In this series, each term is $\frac{1}{4}$ of the previous term, so $r = \frac{1}{4}$ .
Determine Common Ratio: An infinite geometric series converges if the absolute value of the common ratio is less than $1$ (|r| < 1). It diverges if the absolute value of the common ratio is greater than or equal to $1$ ( $|r| \geq 1$ ). $\newline$ In this case, $|r| = |\frac{1}{4}| = 0.25$ , which is less than $1$ .
Check Convergence: Since the absolute value of the common ratio is less than $1$ , the series converges. $\newline$ Therefore, the correct choice is $(A)$ converge.