Can the series sum from n=1 to infinity (sin nx)/n, x
Can the series $\displaystyle\sum_{n=1}^{\infty}\frac{\sin nx}{n}$, $x\in(\varepsilon,2\pi-\varepsilon)$, be differentiated term by term?
Explanation of Method
The original series converges (uniformly, in fact, by Dirichlet's test) on $(\varepsilon,2\pi-\varepsilon)$. But its formal derivative series is $\sum\cos(nx)$, whose terms do not tend to $0$ — so the hypothesis needed for term-by-term differentiation fails, and the technique cannot be applied.
- Confirm the original series converges (by Dirichlet's test).
- Compute the formal derivative series $\sum\cos(nx)$.
- Show this derivative series diverges for every $x$ in the interval.
Step 1 · Convergence of the original series
The partial sums of $\sin(nx)$ are uniformly bounded for $x$ bounded away from multiples of $2\pi$, and $1/n\downarrow0$; by Dirichlet's test, $\sum\sin(nx)/n$ converges (uniformly) on $(\varepsilon,2\pi-\varepsilon)$.
Step 2 · The formal derivative
$u_n'(x)=\dfrac1n\cdot\cos(nx)\cdot n=\cos(nx)$, so the candidate derivative series is $\sum_{n=1}^{\infty}\cos(nx)$.
Step 3 · This series diverges
For $\cos(nx)\to0$ we would need $\lim_{n\to\infty}\cos(nx)=0$, but $\cos(nx)$ oscillates in $[-1,1]$ and never settles to $0$ for any fixed $x\ne2k\pi$ — e.g. at $x=\pi$: $\cos(n\pi)=(-1)^n$, alternating between $\pm1$. By the Divergence Test, $\sum\cos(nx)$ diverges at every $x\in(\varepsilon,2\pi-\varepsilon)$. Since the theorem's hypothesis (uniform convergence of the derivative series) fails, term-by-term differentiation is not valid.
Summary Table
| Series | Behavior |
|---|---|
| $\sum\sin(nx)/n$ | Converges (Dirichlet's test) |
| $\sum\cos(nx)$ (formal derivative) | Diverges everywhere on the interval |
| Term-by-term differentiation | Not valid |
Additional Notes
- The sum itself is the classical sawtooth Fourier series $f(x)=\sum\sin(nx)/n=\dfrac{\pi-x}2$ on $(0,2\pi)$, whose true derivative is the constant $-\tfrac12$ — nothing like the divergent series $\sum\cos(nx)$, confirming that no valid derivative series exists via the naive term-by-term route.
- Contrast directly with Problem 217, where multiplying by an extra $1/n$ (giving $\sum\sin(nx)/n^2$) produces a derivative series $\sum\cos(nx)/n$ that converges uniformly by Dirichlet's test — one extra power of decay is exactly the difference between success and failure.
- This is the Fourier-series echo of Problem 210's factorial/geometric example: both illustrate that uniform convergence of $\sum u_n(x)$ says nothing by itself about the convergence of $\sum u_n'(x)$.
Result
No — the series of derivatives diverges: