Question

For \( x \in (0, \pi) \), let \[ u_n(x) = \frac{\sin(nx)}{\sqrt{n}}, \quad n = 1, 2, 3, \dots \] Then, which of the following is TRUE?

• A. \( \sum_{n=1}^{\infty} u_n(x) \) converges uniformly on \( (0, \pi) \)
• B. \( \sum_{n=1}^{\infty} u_n(x) \) converges uniformly on \( (0, \pi) \)
• C. \( \sum_{n=1}^{\infty} u_n(x) \) converges pointwise but not uniformly on \( (0, \pi) \)
• D. \( \sum_{n=1}^{\infty} u_n(x) \) converges uniformly on every compact subset of \( (0, \pi) \)

Hint:

When checking for uniform convergence, make sure to apply the Weierstrass M-test or other convergence criteria, especially when dealing with trigonometric series. Uniform convergence on compact sets often holds even if it does not hold globally.

Answer 1

The Correct Option is C, D

We are given \( u_n(x) = \frac{\sin(nx)}{\sqrt{n}} \). To determine the convergence of the series, we first check pointwise convergence.

Step 1: Pointwise Convergence
The sequence \( u_n(x) \) converges pointwise because \( \frac{\sin(nx)}{\sqrt{n}} \) tends to zero as \( n \to \infty \) for any fixed \( x \in (0, \pi) \). Therefore, the series converges pointwise.

Step 2: Uniform Convergence
To check for uniform convergence, we use the Weierstrass M-test. Since \( \frac{1}{\sqrt{n}} \) decreases as \( n \) increases, the sum \( \sum_{n=1}^{\infty} \frac{\sin(nx)}{\sqrt{n}} \) does not converge uniformly because the terms do not approach zero uniformly for all \( x \in (0, \pi) \). Therefore, (C) is TRUE.

Step 3: Uniform Convergence on Compact Subsets
The sum \( \sum_{n=1}^{\infty} u_n(x) \) converges uniformly on every compact subset of \( (0, \pi) \) because the series of functions \( u_n(x) \) are continuous and decay sufficiently fast on compact subsets. Hence, (D) is TRUE.

Final Answer
\[ \boxed{(C) \quad \sum_{n=1}^{\infty} u_n(x) \text{ converges pointwise but not uniformly on } (0, \pi)} \] \[ \boxed{(D) \quad \sum_{n=1}^{\infty} u_n(x) \text{ converges uniformly on every compact subset of } (0, \pi)} \]