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Does the infinite geometric series converge or diverge?\newline1+14+116+164+…1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \ldots\newlineChoices:\newline(A) converge\newline(B) diverge

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Convergent and divergent geometric seriesright chevron icon

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Q. Does the infinite geometric series converge or diverge?\newline1+14+116+164+…1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \ldots\newlineChoices:\newline(A) converge\newline(B) diverge
  1. Identify Terms: To determine if the infinite geometric series converges or diverges, we need to identify the first term aa and the common ratio rr of the series.\newlineThe first term a=1a = 1.\newlineThe common ratio rr is the factor by which each term is multiplied to get the next term. In this series, each term is 14\frac{1}{4} of the previous term, so r=14r = \frac{1}{4}.
  2. Determine Common Ratio: An infinite geometric series converges if the absolute value of the common ratio is less than 11 (∣r∣<1|r| < 1). It diverges if the absolute value of the common ratio is greater than or equal to 11 (∣r∣≥1|r| \geq 1).\newlineIn this case, ∣r∣=∣14∣=0.25|r| = |\frac{1}{4}| = 0.25, which is less than 11.
  3. Check Convergence: Since the absolute value of the common ratio is less than 11, the series converges.\newlineTherefore, the correct choice is (A)(A) converge.

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