Question

Which one of the following series is divergent?

• A. \( \sum_{n=1}^{\infty} \frac{\sin^2 \left( \frac{1}{n} \right)}{n^2} \)
• B. \( \sum_{n=1}^{\infty} \frac{1}{n \log n} \)
• C. \( \sum_{n=1}^{\infty} \frac{1}{n^3} \)
• D. \( \sum_{n=1}^{\infty} \frac{1}{n \tan^{-1} n} \)

Hint:

To determine if a series converges or diverges, apply the integral test or comparison test, particularly for series involving logarithmic terms.

Answer 1

The Correct Option is B

Step 1: Analyze the series.
We are asked to identify the divergent series. Let's analyze each series: (A) \( \sum_{n=1}^{\infty} \frac{\sin^2 \left( \frac{1}{n} \right)}{n^2} \): This series converges by the comparison test, as \( \sin \left( \frac{1}{n} \right) \approx \frac{1}{n} \) for large \( n \).
(B) \( \sum_{n=1}^{\infty} \frac{1}{n \log n} \): This series diverges by the integral test, since the integral of \( \frac{1}{x \log x} \) diverges.
(C) \( \sum_{n=1}^{\infty} \frac{1}{n^3} \): This series converges as it is a p-series with \( p = 3>1 \).
(D) \( \sum_{n=1}^{\infty} \frac{1}{n \tan^{-1} n} \): This series converges because \( \tan^{-1} n \) grows asymptotically like \( n \), so the series behaves like a convergent p-series.
Step 2: Conclusion.
Thus, the divergent series is \( \boxed{(B)} \).