Does the infinite geometric series converge or diverge? 1 + 7/3 + 49/9 + 343/27 + Choices: (A)converge (B)diverge

Identify Common Ratio: Identify the common ratio of the geometric series by dividing a term by its preceding term. We can take the second term (7/3) and divide it by the first term (1): Common Ratio (r) = (7/3) / 1 = 7/3Find Absolute Value: Find the absolute value of the common ratio. |r| = |7/3| |r| = 7/3Determine Convergence: Determine if the given geometric series converges or diverges. A geometric series converges if the absolute value of the common ratio is less than 1. Since |r| = 7/3 > 1, the series diverges.

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Does the infinite geometric series converge or diverge? $\newline$ $1 + \frac{7}{3} + \frac{49}{9} + \frac{343}{27} + \ldots$ $\newline$ 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{7}{3} + \frac{49}{9} + \frac{343}{27} + \ldots

\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. $\newline$ We can take the second term $(\frac{7}{3})$ and divide it by the first term $(1)$ : $\newline$ Common Ratio $(r) = \frac{\frac{7}{3}}{1} = \frac{7}{3}$
Find Absolute Value: Find the absolute value of the common ratio. $\newline$ |r| = |\frac{$7$}{$3$}|\(\newline|r| = \frac{ $7$ }{ $3$ } $\newline$ )
Determine Convergence: Determine if the given geometric series converges or diverges. $\newline$ A geometric series converges if the absolute value of the common ratio is less than $1$ . $\newline$ Since |r| = \frac{7}{3} > 1, the series diverges.