Question:

Let \( (a_n) \) be a sequence of positive real numbers. The series \[ \sum_{n=1}^{\infty} a_n^2 \quad \text{converges if the series} \]

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For sequences of positive real numbers, if \( \sum a_n \) converges, then \( \sum a_n^2 \) will also converge.
Updated On: Dec 11, 2025
  • \( \sum_{n=1}^{\infty} a_n \) converges
  • \( \sum_{n=1}^{\infty} \frac{1}{n} a_n \) converges
  • \( \sum_{n=1}^{\infty} \frac{1}{n^2} a_n \) converges
  • \( \sum_{n=1}^{\infty} \frac{a_n}{a_{n+1}} \) converges
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The Correct Option is C

Solution and Explanation

Step 1: Understanding the convergence of the series.
To determine convergence, we apply the comparison test. If \( \sum a_n \) converges, then \( \sum a_n^2 \) also converges. This is because \( a_n^2 \) is a smaller term than \( a_n \) for all sufficiently small \( a_n \).
Step 2: Conclusion.
Thus, the correct answer is \( \boxed{(A)} \).
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