Does the infinite geometric series converge or diverge? 1 + 5/3 + 25/9 + 125/27 + Choices: (A)converge (B)diverge

Find Common Ratio: To determine if the series converges or diverges, we need to find the common ratio (r) of the geometric series. We can do this by dividing a term in the series by the term before it. Let's take the second term (5/3) and divide it by the first term (1): r = (5/3) / 1 = 5/3Check Absolute Value: Now that we have the common ratio r = 5/3, we need to check its absolute value to determine if the series converges or diverges. The absolute value of r is |r| = |5/3| = 5/3.Determine Convergence: For an infinite geometric series to converge, the absolute value of the common ratio must be less than 1. In this case, |r| = 5/3, which is greater than 1. Since |r| > 1, the series diverges.

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

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Math Problems

Algebra 2

Convergent and divergent geometric series

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Q. Does the infinite geometric series converge or diverge?

\newline

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

\newline

Choices:

\newline

(A)converge

\newline

(B)diverge

Find Common Ratio: To determine if the series converges or diverges, we need to find the common ratio $r$ of the geometric series. We can do this by dividing a term in the series by the term before it. $\newline$ Let's take the second term $\frac{5}{3}$ and divide it by the first term $1$ : $\newline$ $r = \frac{5}{3} / 1 = \frac{5}{3}$
Check Absolute Value: Now that we have the common ratio $r = \frac{5}{3}$ , we need to check its absolute value to determine if the series converges or diverges. $\newline$ The absolute value of $r$ is $|r| = \left|\frac{5}{3}\right| = \frac{5}{3}$ .
Determine Convergence: For an infinite geometric series to converge, the absolute value of the common ratio must be less than $1$ . In this case, $|r| = \frac{5}{3}$ , which is greater than $1$ . $\newline$ Since |r| > 1, the series diverges.