Test for pointwise and uniform convergence
Test for pointwise and uniform convergence of $u_n(x)=\dfrac{\sin nx}n$ on an arbitrary interval $[a,b]$.
Explanation of Method
The universal bound $|\sin(nx)|\le1$ immediately squeezes $u_n(x)$ to $0$ at every point, and — crucially — this same bound is completely independent of $x$, so the supremum over the whole interval is trapped between $0$ and $1/n$ regardless of how large or small $[a,b]$ is. This gives uniform convergence essentially for free, without needing any calculus-based optimization.
- Find the pointwise limit via the squeeze theorem.
- Bound the supremum of $|u_n(x)|$ over $[a,b]$ using $|\sin(nx)|\le1$.
- Confirm the bound is sharp (attained, for large $n$) and still tends to $0$.
Step 1 · Pointwise limit
$|\sin(nx)/n|\le1/n\to0$ for every fixed $x$, so by the squeeze theorem $u_n(x)\to0$ pointwise.
Step 2 · Bound the supremum
$\sup_{x\in[a,b]}|u_n(x)|\le\dfrac1n$ for every $n$, since $|\sin(nx)|\le1$ uniformly in $x$.
Step 3 · Sharpness and conclusion
For $n$ large enough that $\frac{2\pi}n<b-a$ (so $\sin(nx)$ completes at least one full period on $[a,b]$), the bound is attained: $\sup_{x\in[a,b]}|\sin(nx)|=1$ exactly, so $\sup_{x\in[a,b]}|u_n(x)|=1/n$ for all sufficiently large $n$. Either way, $$ \lim_{n\to\infty}\sup_{x\in[a,b]}|u_n(x)-0| = \lim_{n\to\infty}\frac1n = 0. $$ Uniform convergence holds.
Summary Table
| Quantity | Value |
|---|---|
| Pointwise limit | $u(x)\equiv0$ |
| $\sup_{[a,b]}|u_n-u|$ | $\le1/n$ always; $=1/n$ for large $n$ |
| Convergence | Uniform, rate $O(1/n)$ |
Additional Notes
- The oscillation frequency $n$ inside $\sin(nx)$ plays no role in the convergence conclusion — only the outer factor $1/n$ matters, since the sine's amplitude is always capped at $1$ regardless of how fast it oscillates. This is a useful contrast with a series like $\sum\sin(nx)/n^a$ (Problem 133), where the interplay between oscillation and decay rate genuinely matters for series convergence.
- This is the cleanest possible example of a "dampened oscillation" sequence converging uniformly: the argument requires no calculus (no derivative, no critical point), unlike Problems 177 or 178, since the elementary bound $|\sin(nx)|\le1$ does all the work.
- The result holds for any bounded interval $[a,b]$, however large — a genuinely global statement, matching the strength of Problem 179's uniform convergence on all of $\mathbb R$, though by a much more elementary argument here.
Result
Converges uniformly to $0$ on any interval $[a,b]$: