Uniform convergence of series $\sum\limits_{n=2}^\infty\frac{\sin n x}{n\log n}$

Using Dirichlet series test I proved that the series $\displaystyle\sum\limits_{n=2}^\infty\frac{\sin n x}{n\log n}$ converges for all $x\in\mathbb{R}$.

How to determine whether this series converges uniformly on $\mathbb{R}$?

Solution 1:

Let $\left\{a_n\right\}$ be a decreasing sequence of real numbers such that $n\cdot a_n\to 0$. Then the series $\sum_{n\geqslant 2}a_n\sin (nx)$ is uniformly convergent on $\mathbb R$.

Thanks to Abel transform, we can show that the convergence is uniform on $[\delta,2\pi-\delta]$ for all $\delta>0$. Since the functions are odd, we only have to prove the uniform convergence on $[0,\delta]$. Put $M_n:=\sup_{k\geqslant n}ka_k$, and $R_n(x)=\sum_{k=n}^{\infty}a_k\sin (kx)$. Fix $x\neq 0$ and $N$ such that $\frac 1N\leqslant x<\frac 1{N-1}$. Put for $n\lt N$: $$A_n(x)=\sum_{k=n}^{N-1}a_k\sin kx\mbox{ and }B_n(x):=\sum_{k=n}^{+\infty}a_k\sin (kx),$$ and for $n\geq N$, $A_n(x):=0$.

Since $|\sin t|\leqslant t$ for $t\geq 0$ we have $$|A_n(x)|\leqslant \sum_{k=n}^{N-1}a_kkx\leqslant M_nx(N-n)\leqslant \frac{N-n}{N-1}M_n,$$ so $|A_n(x)|\leqslant M_n$.

If $N> n$, we have after writing $D_k=\sum_{j=0}^k\sin jx$, $|D_k(x)|\leqslant \frac cx$ on $(0,\delta]$ for some constant $c$. Indeed, we have $|D_k(x)|\leqslant \frac 1{\sqrt{2(1-\cos x)}}$ and $\cos x=1-\frac{x^2}2(1+\xi)$ where $|\xi|\leqslant \frac 12$ so $2(1-\cos x)\geqslant \frac{x^2}2$ and $|D_k(x)\leqslant \frac{\sqrt 2}x$. Therefore $$|B_n(x)|\leqslant \frac{\sqrt 2}x\sum_{k=N}^{+\infty}(a_k-a_{k+1})+a_N\frac{\sqrt 2}x=\frac{2\sqrt 2}xa_N\leqslant 2\sqrt 2 Na_N\leqslant 2\sqrt 2M_n.$$ We get the same bound if $N\leqslant n$. Finally $|R_n(x)|\leqslant (2\sqrt 2+1)M_n$ for all $0\leqslant x\leqslant \delta$, so the convergence is uniform on $\mathbb R$.

Added latter: it's an example of a Fourier series which is uniformly convergent on the real line, but not absolutely convergent at any point of $(0,2\pi)$. Indeed take $x\in (0,2\pi)$. Since $|\sin(nx)|\geqslant \sin^2(nx)$, we would have the convergence of $\sum_{n\geqslant 2}\frac{\sin^2(nx)}{n\log n}$. We have $\sin ^2(nx)= \frac 1{-4}(e^{inx}-e^{-inx})^2=-\frac 14 (e^{2inx}+e^{-2inx}-2)=\frac 12-\frac 12\cos (2nx)$ and an Abel transform shows that the series $\sum_{n\geqslant 2}\frac{\cos(2nx)}{n\log n}$ is convergent. So the series $\sum_{n\geqslant 2}\frac 1{n\log n}$ would be convergent, which is not the case as the integral test shows.