Does the infinite geometric series converge or diverge? 1 + 4 + 16 + 64 + 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 found by dividing any term in the series by the preceding term. For this series, we can use the second term (4) and the first term (1) to find the common ratio. r = 4 / 1 = 4Check Convergence Criterion: An infinite geometric series converges if the absolute value of the common ratio is less than 1 (|r| < 1). Since we found that r = 4, we can see that |r| = |4| = 4, which is not less than 1. Therefore, the series does not meet the convergence criterion.Series Diverges: Since the common ratio is greater than 1, the series diverges. The terms of the series will continue to grow larger without bound, and the sum of the series cannot approach a finite number.

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Does the infinite geometric series converge or diverge? $\newline$ $1 + 4 + 16 + 64 + \dots$ $\newline$ 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 + 4 + 16 + 64 + \dots

\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 found by dividing any term in the series by the preceding term. $\newline$ For this series, we can use the second term $4$ and the first term $1$ to find the common ratio. $\newline$ $r = \frac{4}{1} = 4$
Check Convergence Criterion: An infinite geometric series converges if the absolute value of the common ratio is less than $1$ (|r| < 1). $\newline$ Since we found that $r = 4$ , we can see that $|r| = |4| = 4$ , which is not less than $1$ . $\newline$ Therefore, the series does not meet the convergence criterion.
Series Diverges: Since the common ratio is greater than $1$ , the series diverges. $\newline$ The terms of the series will continue to grow larger without bound, and the sum of the series cannot approach a finite number.