# Does $\sum_{n=1}^{\infty} \frac{\sin^2(nx)}{n}$ converge uniformly?

Does $\sum_{n=1}^{\infty} \frac{\sin^2(nx)}{n}$ converge uniformly?

I cannot apply the Weierstrass M-test to this series, but I know $\sum_{n=1}^{\infty} \frac{\sin(nx)}{n}$ does not converge uniformly, are these related? Any help is appreciated.

## 2 Answers

Note that $\sin^2(nx)=\frac12-\frac12\cos(2nx)$. Then,

$$\sum_{n=1}^N \frac{\sin^2(nx)}{n}=\frac12\sum_{n=1}^N\frac{1}{n}-\frac12\sum_{n=1}^N\frac{\cos(2nx)}{n} \tag 1$$

The second term on the right-hand side of $(1)$ converges (use the Dirichlet test) for all $x$, but the second term, the harmonic series, diverges. Therefore, if $x\ne n\pi$, then the series of interest diverges.

Additionally, we note that for $0<r_1<x<2\pi - r_2$ and $r_2>0$ the series $\sum_{n\ge 1}\frac{\sin(xn)}{n}$ converges uniformly (use the Dirichlet test).

If the series converged uniformly, it would be uniformly Cauchy. This would imply that for large even $N,$

$$\left |\sum_{n=N/2}^{N}\frac{\sin^2(n/N)}{n}\right| < 1.$$

But that sum is at least $(N/2)\sin(1/2),$ which can be made as large as desired, contradiction.