Answer: Conditionally Convergent Series of Functions: Pointwise but Not Uniform

Key Idea / Intuition

The series converges pointwise by the Dirichlet test: the partial sums of $\sin(nx)$ are bounded (they satisfy a telescoping geometric sum estimate) and $1/n \searrow 0$ . But uniform convergence fails because near $x = 0$ the terms $\sin(nx)/n$ don't die uniformly — you can always find an $x$ small enough that the partial sums are large. In short: pointwise convergence can hold everywhere while uniform convergence fails because the "trouble spot" chases $x \to 0$ .

Formal Proof / Solution

Part (a): Pointwise Convergence

Fix $x \in (0, 2\pi)$ . We apply the Dirichlet test for series: if $b_n = 1/n \searrow 0$ and the partial sums $A_N(x) = \sum_{n=1}^N (-1)^n \sin(nx)$ are bounded in $N$ , then the series converges.

We bound the partial sums using the identity $\sum_{n=1}^N (-1)^n \sin(nx) = \text{Im}\!\left(\sum_{n=1}^N (-e^{ix})^n\right) = \text{Im}\!\left(\frac{-e^{ix}(1-(-e^{ix})^N)}{1+e^{ix}}\right).$

Since $x \in (0, 2\pi)$ , we have $e^{ix} \neq -1$ , so $|1 + e^{ix}| = 2|\cos(x/2)| > 0$ . Therefore $|A_N(x)| \leq \frac{2}{|1 + e^{ix}|} = \frac{1}{|\cos(x/2)|} < \infty.$

This bound is finite for each fixed $x \in (0, 2\pi)$ , so by the Dirichlet test, the series converges pointwise. $\checkmark$

Part (b): Convergence Is Not Uniform

Suppose for contradiction the convergence were uniform on $(0, 2\pi)$ . Then the partial sums $S_N(x) = \sum_{n=1}^{N} \frac{(-1)^n}{n}\sin(nx)$ satisfy $\sup_{x \in (0,2\pi)} |f(x) - S_N(x)| \to 0$ .

In particular, the tail $R_N(x) = f(x) - S_N(x)$ would go to zero uniformly, so in particular $S_N(x)$ would be uniformly Cauchy, meaning

$\sup_{x \in (0,2\pi)} \left|\sum_{n=N+1}^{M} \frac{(-1)^n}{n}\sin(nx)\right| \to 0 \quad \text{as } N \to \infty.$

Now consider the single term remainder: take $M = N+1$ , so we need $\sup_{x \in (0,2\pi)} \left|\frac{(-1)^{N+1}}{N+1}\sin((N+1)x)\right| = \frac{1}{N+1}\sup_{x} |\sin((N+1)x)| = \frac{1}{N+1} \to 0.$

This part is fine. The real issue is more subtle — let us use a necessary condition for uniform convergence.

Key test: If $\sum f_n$ converges uniformly on $(0,2\pi)$ , then $\sup_x |f_n(x)| \to 0$ . Here $f_n(x) = \frac{(-1)^n}{n}\sin(nx)$ , and $\sup_{x \in (0,2\pi)} \left|\frac{\sin(nx)}{n}\right| = \frac{1}{n} \to 0.$ This is consistent, so this test does not immediately give a contradiction. We need a sharper argument.

Better approach — look at $S_N$ near $x = 0$ .

It is a classical fact (Fourier series of a sawtooth) that the series $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\sin(nx)$ represents $x/2$ on $(-\pi, \pi)$ . Rearranging signs,

$\sum_{n=1}^{\infty} \frac{(-1)^n}{n}\sin(nx) = -\frac{x}{2}, \quad x \in (0, 2\pi).$

So $f(x) = -x/2$ , which is continuous on $(0,2\pi)$ but has a jump discontinuity at $0$ and $2\pi$ .

Now, each partial sum $S_N(x)$ is a finite sum of continuous functions, hence continuous and bounded on $[0, 2\pi]$ with $S_N(0) = 0$ for all $N$ .

If $S_N \to f = -x/2$ uniformly on $(0, 2\pi)$ , then $f$ would extend to a continuous function on $[0, 2\pi]$ (as the uniform limit of continuous functions). But $f(x) = -x/2 \to 0$ as $x \to 0^+$ while we also have $S_N(0) = 0$ for all $N$ . So far no contradiction.

The actual non-uniformity is near $x = 2\pi$ : As $x \to 2\pi^-$ , $f(x) = -x/2 \to -\pi$ , but $S_N(2\pi) = 0$ for all $N$ (since $\sin(2\pi n) = 0$ ). So

$\sup_{x \in (0, 2\pi)} |f(x) - S_N(x)| \geq \lim_{x \to 2\pi^-} |{-x/2} - S_N(x)| = |-\pi - 0| = \pi.$

This shows $\sup |f - S_N| \geq \pi > 0$ for all $N$ , so convergence is not uniform. $\blacksquare$

Summary

| | What holds | |---|---| | Pointwise convergence on $(0,2\pi)$ | ✓ (Dirichlet test) | | Uniform convergence on $(0,2\pi)$ | ✗ (limit function has a boundary jump) |