Does the infinite geometric series converge or diverge? 1 + 8/3 + 64/9 + 512/27 + Choices: (A)converge (B)diverge

Identify common ratio: Identify the common ratio of the geometric series by dividing a term by its preceding term. Using the second term (8/3) and the first term (1), we calculate the common ratio (r) as follows: r = (8/3) / 1 = 8/3Calculate absolute value: Find the absolute value of the common ratio. |r| = |8/3| Since 8/3 is greater than 1, its absolute value is also greater than 1.Determine convergence or divergence: Determine if the given geometric series converges or diverges based on the absolute value of the common ratio. Since |r| = 8/3 > 1, the series diverges according to the convergence test for geometric series, which states that a geometric series converges if and only if |r| < 1.

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Does the infinite geometric series converge or diverge? $\newline$ $1 + \frac{8}{3} + \frac{64}{9} + \frac{512}{27} + \dots$ $\newline$ 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{8}{3} + \frac{64}{9} + \frac{512}{27} + \dots

\newline

Choices:

\newline

(A)converge

\newline

(B)diverge

Identify common ratio: Identify the common ratio of the geometric series by dividing a term by its preceding term. $\newline$ Using the second term $(\frac{8}{3})$ and the first term $(1)$ , we calculate the common ratio $(r)$ as follows: $\newline$ $r = (\frac{8}{3}) / 1 = \frac{8}{3}$
Calculate absolute value: Find the absolute value of the common ratio. $|r| = |\frac{8}{3}|$ Since $\frac{8}{3}$ is greater than $1$ , its absolute value is also greater than $1$ .
Determine convergence or divergence: Determine if the given geometric series converges or diverges based on the absolute value of the common ratio. $\newline$ Since |r| = \frac{8}{3} > 1, the series diverges according to the convergence test for geometric series, which states that a geometric series converges if and only if |r| < 1.