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.
First, because is an alternating series, we start with the alternating series test: if $(a_n)\to 0$ and $|a_n|$ decreases then $\sum(-1)^{n}a_n$ converges.
We can see that $\log\left(\frac{n+1}{n}\right)\to 0$ because $\left(\frac{n+1}{n}\right)\to 1$ and $\left|\log\left(\frac{n+1}{n}\right)\right|$ decreases because logarithm is a strict increasing function and $\frac{n+1}{n}$ obviously decreases. Then the series, at least, is conditionally convergent.
To test absolute convergence we can see that we can write
$$\sum_{n=1}^{\infty}\log\left(\frac{n+1}{n}\right)=\sum_{n=1}^{\infty}\log(n+1)-\log(n)$$
that is a telescoping sum. And because $\sum_{n=1}^{\infty}a_n=\lim_{N\to\infty}\sum_{n=1}^{N}a_n$ then
$$\sum_{n=1}^{\infty}\log(n+1)-\log(n)=\lim_{N\to\infty}\sum_{n=1}^{N}\log(n+1)-\log(n)=\lim_{N\to\infty}\log(N+1)$$
what is divergent, so the series is conditionally convergent.
Best Answer
$$ \frac{\sum_{k=1}^{n} \frac{1}{k}}{(\log{n})^2} \ \ \lor \ \ \frac{\sum_{k=1}^{n+1} \frac{1}{k}}{(\log{(n+1)})^2}$$
$$ \left(\sum_{k=1}^{n} \frac{1}{k}\right) \log^2(n+1) - \left(\sum_{k=1}^{n+1} \frac{1}{k}\right) \log^2(n) \ \ \lor \ \ 0$$
$$ \left(\sum_{k=1}^{n} \frac{1}{k}\right) \left[\log^2(n+1) - \log^2(n)\right] - \frac{1}{n+1} \log^2(n) > \\ \{\log^2(n+1) - \log^2(n) > 0\ \ \text{for} \ \ n > 1 \} \\ \log(n+1) \left[\log^2(n+1) - \log^2(n)\right] - \frac{1}{n+1} \log^2(n) > 0 $$
Last inequality flows from Wolfram, but I'm not sure, how to prove it strictly. Any help with it?