Question

If the series \(\sum_{n=1}^{\infty} a_n\) converges absolutely, then which of the following series diverges?

• A. \(\sum_{n=1}^{\infty} |a_{2n}|\)
• B. \(\sum_{n=1}^{\infty} \frac{a_n + a_{n+1}}{2}\)
• C. \(\sum_{n=1}^{\infty} (a_n)^3\)
• D. \(\sum_{n=2}^{\infty} \left(\frac{1}{(\ln n)^2} + a_n\right)\)

Hint:

When a series converges absolutely, any finite manipulation or power of its terms also converges. Adding a divergent part like \(\frac{1}{(\ln n)^2}\) leads to divergence.

Answer 1

The Correct Option is D

Step 1: Recall property of absolute convergence.
If \(\sum a_n\) converges absolutely, then \(\sum |a_n|\) converges, and so do all related series where \(a_n\) is replaced by powers or linear combinations (like \(a_n^3\), \(\frac{a_n + a_{n+1}}{2}\), etc.).
Step 2: Analyze each option.
(A) \(\sum |a_{2n}|\): This is a subseries of \(\sum |a_n|\), so it converges.
(B) \(\sum \frac{a_n + a_{n+1}}{2}\): This converges because both \(\sum a_n\) and its shift \(\sum a_{n+1}\) converge.
(C) \(\sum (a_n)^3\): Since \(a_n \to 0\) and \(|a_n|^3<|a_n|\), this also converges absolutely.
(D) \(\sum \left(\frac{1}{(\ln n)^2} + a_n\right)\): The term \(\sum \frac{1}{(\ln n)^2}\) diverges because \(\frac{1}{(\ln n)^2}\) does not decrease rapidly enough for convergence (it behaves similarly to the harmonic series).
Step 3: Conclusion.
Hence, option (D) diverges. Final Answer: \[ \boxed{(D) \sum_{n=2}^{\infty} \left(\frac{1}{(\ln n)^2} + a_n\right)} \]