Does the infinite geometric series converge or diverge? 1 + 3/5 + 9/25 + 27/125 + 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. Looking at the series, we can see that each term is multiplied by 3/5 to get the next term. So, the common ratio r = 3/5.Apply Sum Formula: Now, we need to apply the formula for the sum of an infinite geometric series, which is S = a / (1 - r), where S is the sum of the series, a is the first term, and r is the common ratio. The series converges if the absolute value of r is less than 1 (|r| < 1). In this case, |r| = |3/5| = 3/5, which is less than 1.Check Convergence: Since the absolute value of the common ratio is less than 1, the infinite geometric series converges. Therefore, the correct choice is (A) converge.

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Does the infinite geometric series converge or diverge? $\newline$ $1 + \frac{3}{5} + \frac{9}{25} + \frac{27}{125} + \ldots$ $\newline$ Choices: $\newline$ (A) converge $\newline$ (B) diverge $\newline$

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Precalculus

Convergent and divergent geometric series

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Q. Does the infinite geometric series converge or diverge?

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1 + \frac{3}{5} + \frac{9}{25} + \frac{27}{125} + \ldots

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Choices:

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(A) converge

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(B) diverge

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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. $\newline$ Looking at the series, we can see that each term is multiplied by $\frac{3}{5}$ to get the next term. $\newline$ So, the common ratio $r = \frac{3}{5}$ .
Apply Sum Formula: Now, we need to apply the formula for the sum of an infinite geometric series, which is $S = \frac{a}{1 - r}$ , where $S$ is the sum of the series, $a$ is the first term, and $r$ is the common ratio. $\newline$ The series converges if the absolute value of $r$ is less than $1$ (|r| < 1). $\newline$ In this case, $|r| = \left|\frac{3}{5}\right| = \frac{3}{5}$ , which is less than $1$ .
Check Convergence: Since the absolute value of the common ratio is less than $1$ , the infinite geometric series converges. $\newline$ Therefore, the correct choice is $(A)$ converge.