Can the series sum from n=1 to infinity (sin nx)/(n²), x
Can the series $\displaystyle\sum_{n=1}^{\infty}\frac{\sin nx}{n^2}$, $x\in(\varepsilon,2\pi-\varepsilon)$, be differentiated term by term?
Explanation of Method
The original series converges uniformly on $\mathbb R$ by the Weierstrass M-test. The formal derivative series is $\sum\cos(nx)/n$, which — unlike Problem 216's $\sum\cos(nx)$ — converges uniformly on $(\varepsilon,2\pi-\varepsilon)$ by Dirichlet's test, since the partial sums of $\cos(nx)$ are uniformly bounded away from $x=2k\pi$ and $1/n\downarrow0$. With both hypotheses satisfied, term-by-term differentiation is valid.
- Confirm uniform convergence of the original series via the Weierstrass M-test.
- Compute the formal derivative series $\sum\cos(nx)/n$.
- Verify uniform convergence of the derivative series via Dirichlet's test.
Step 1 · Uniform convergence of the original series
Since $|\sin nx|\le1$: $|\sin(nx)/n^2|\le1/n^2$, and $\sum1/n^2=\pi^2/6<\infty$. By the Weierstrass M-test, the series converges uniformly and absolutely on $\mathbb R$, in particular on $(\varepsilon,2\pi-\varepsilon)$.
Step 2 · The formal derivative
$u_n'(x)=\dfrac1{n^2}\cdot\cos(nx)\cdot n=\dfrac{\cos nx}n$, so the candidate derivative series is $\sum_{n=1}^{\infty}\dfrac{\cos nx}n$.
Step 3 · Uniform convergence of the derivative series, via Dirichlet's test
For $x\in(\varepsilon,2\pi-\varepsilon)$, $0<\varepsilon<\pi$: the partial sums $S_N(x)=\sum_{k=1}^{N}\cos(kx)=\dfrac{\sin((N+\frac12)x)}{2\sin(x/2)}-\dfrac12$ satisfy $|S_N(x)|\le\dfrac1{2\sin(\varepsilon/2)}+\dfrac12=:M$, a bound independent of $N$ and $x$ (since $\sin(x/2)\ge\sin(\varepsilon/2)>0$ throughout the interval). The sequence $1/n$ is monotone decreasing to $0$. By Dirichlet's test for uniform convergence, $\sum\cos(nx)/n$ converges uniformly on $(\varepsilon,2\pi-\varepsilon)$.
Since each $u_n(x)=\sin(nx)/n^2$ is differentiable, the original series converges, and the derivative series converges uniformly, term-by-term differentiation is valid: $$ \frac{d}{dx}\left[\sum_{n=1}^{\infty}\frac{\sin nx}{n^2}\right] = \sum_{n=1}^{\infty}\frac{\cos nx}n. $$
Summary Table
| Series | Behavior |
|---|---|
| $\sum\sin(nx)/n^2$ | Converges uniformly (M-test) |
| $\sum\cos(nx)/n$ (derivative) | Converges uniformly on $(\varepsilon,2\pi-\varepsilon)$ (Dirichlet's test) |
| Term-by-term differentiation | Valid |
Additional Notes
- This is the mirror image of Problem 216: replacing the coefficient $1/n$ by $1/n^2$ is exactly the extra decay needed to push the formal derivative series from divergent ($\sum\cos nx$) to uniformly convergent ($\sum\cos(nx)/n$, itself only convergent via Dirichlet's test, not absolutely).
- Dirichlet's uniform-convergence test used here is the exact tool behind Problem 205's Abel's-test proof — both rely on bounding partial sums of an oscillating factor uniformly in the parameter, then pairing that with a monotone decreasing sequence.
- Iterating this differentiation once more (formally differentiating $\sum\cos(nx)/n$ term by term) would reproduce Problem 216's divergent series $-\sum\sin(nx)$-type behavior — so the sum here is differentiable exactly once via this route, not twice.
Result
Yes — valid by Dirichlet's uniform-convergence test on the derivative series: