Question

Consider the following statements:
(I) If \(a_n \ge 0\) for each \(n \in \mathbb{N}\) and the series \(\sum_{n=1}^\infty a_n\) converges, then the series \(\sum_{n=1}^\infty \dfrac{\sqrt{a_n}}{n^p}\) converges for \(p > 1/2\).
(II) The series \(\sum_{n=1}^\infty (-1)^n \sin(1/n)\) is conditionally convergent.
Which one of the options given below is correct?

• A. Only (I) is true
• B. Only (II) is true
• C. Both (I) and (II) are true
• D. Neither (I) nor (II) is true

Hint:

Use the Cauchy-Schwarz inequality for (I) and the Leibniz test with limit comparison for (II).

Answer 1

The Correct Option is C

Step 1: Set up statement (I) with the Cauchy-Schwarz inequality. Write \(\dfrac{\sqrt{a_n}}{n^p} = \sqrt{a_n}\cdot\dfrac{1}{n^p}\). For any \(N\), \[\sum_{n=1}^{N}\sqrt{a_n}\cdot\frac{1}{n^p} \le \left(\sum_{n=1}^{N}a_n\right)^{1/2}\left(\sum_{n=1}^{N}\frac{1}{n^{2p}}\right)^{1/2}.\]

Step 2: Check both factors converge. Since \(\sum a_n\) converges, the first factor stays bounded. Since \(p > 1/2\) gives \(2p > 1\), the p-series \(\sum \dfrac{1}{n^{2p}}\) converges, so the second factor is also bounded. The partial sums of \(\sum \dfrac{\sqrt{a_n}}{n^p}\) are therefore bounded, and because every term is nonnegative, the series converges. Statement (I) is true.

Step 3: Set up statement (II). Let \(b_n = \sin(1/n)\). For every \(n \ge 1\), \(0 < 1/n \le 1 < \pi/2\), and \(\sin\) is increasing on \((0,\pi/2)\), so \(b_n\) is a decreasing sequence with \(b_n \to 0\).

Step 4: Apply the alternating series test. Since \(b_n \downarrow 0\), the Leibniz test guarantees \(\sum(-1)^n b_n = \sum (-1)^n\sin(1/n)\) converges.

Step 5: Check absolute convergence. Since \(\lim_{n\to\infty}\dfrac{\sin(1/n)}{1/n} = 1\), the limit comparison test shows \(\sum \sin(1/n)\) diverges exactly like the harmonic series \(\sum 1/n\). So \(\sum|(-1)^n\sin(1/n)|\) diverges.

Step 6: Conclude. The series converges but not absolutely, so it is conditionally convergent. Statement (II) is true. Both statements hold. \[\boxed{\text{Both (I) and (II) are true}}\]