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

To determine the nature of convergence for the series \sum_{n=1}^{\infty} u_n(x) where u_n(x) = \frac{\sin(nx)}{\sqrt{n}} , we analyze both pointwise and uniform convergence over the interval (0, \pi) .

Pointwise Convergence:

• For each fixed x \in (0, \pi) , we consider the sequence of terms \frac{\sin(nx)}{\sqrt{n}} .
• The sequence \sin(nx) oscillates between -1 and 1, thus \left| \frac{\sin(nx)}{\sqrt{n}} \right| \leq \frac{1}{\sqrt{n}} .
• The series \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} is a divergent p-series with p = \frac{1}{2} .
• Therefore, by the comparison test, the series \sum_{n=1}^{\infty} \left| \frac{\sin(nx)}{\sqrt{n}} \right| diverges.
• However, individual terms go to zero, so the series \sum_{n=1}^{\infty} \frac{\sin(nx)}{\sqrt{n}} converges pointwise to zero for each fixed x .

Uniform Convergence:

• For uniform convergence over (0, \pi) , we must have \lim_{n \to \infty} \sup_{x \in (0, \pi)} \left| \frac{\sin(nx)}{\sqrt{n}} \right| = 0 .
• Since \sin(nx) oscillates indefinitely, the supremum is approximately 1 for large n .
• This implies that \sup_{x \in (0, \pi)} \left| \frac{\sin(nx)}{\sqrt{n}} \right| = \frac{1}{\sqrt{n}} , which does not tend to zero.
• Hence, the series does not converge uniformly on (0, \pi) .

Convergence on Compact Subsets:

• Consider a compact subset [a, b] \subset (0, \pi) .
• In compact intervals, the function values are restricted, making the oscillations less significant.
• By the Weierstrass M-test, where M_n = \frac{1}{\sqrt{n}} is a convergent series for bounded n , the series \sum_{n=1}^{\infty} \frac{\sin(nx)}{\sqrt{n}} converges uniformly on every compact subset of (0, \pi) .

Thus, the correct answer is that the series \sum_{n=1}^{\infty} u_n(x) converges pointwise but not uniformly on (0, \pi) , and it converges uniformly on every compact subset of (0, \pi) .