Does the infinite geometric series converge or diverge? 1 + 1/3 + 1/9 + 1/27 + 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 the factor by which each term is multiplied to get the next term. In this series, the second term is 1/3 and the first term is 1, so the common ratio r is (1/3) / 1 = 1/3.Check Convergence Criteria: An infinite geometric series converges if the absolute value of the common ratio |r| is less than 1. In this case, |r| = |1/3| = 1/3, which is less than 1.Conclusion: Since the absolute value of the common ratio is less than 1, the series converges. Therefore, the correct choice is (A) converge.

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

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Precalculus

Convergent and divergent geometric series

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Q. Does the infinite geometric series converge or diverge?

\newline

1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{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 the factor by which each term is multiplied to get the next term. $\newline$ In this series, the second term is $\frac{1}{3}$ and the first term is $1$ , so the common ratio $r$ is $\left(\frac{1}{3}\right) / 1 = \frac{1}{3}$ .
Check Convergence Criteria: An infinite geometric series converges if the absolute value of the common ratio $|r|$ is less than $1$ . $\newline$ In this case, $|r| = |\frac{1}{3}| = \frac{1}{3}$ , which is less than $1$ .
Conclusion: Since the absolute value of the common ratio is less than $1$ , the series converges. $\newline$ Therefore, the correct choice is $(A)$ converge.