Uniform convergence of series

[Tolaso J Kos]

Examine if the series

$$ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\sin (\pi n x) , \; x \in \mathbb{R}$$

converges uniformly.

[Grigorios Kostakos]

The series \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\sin (\pi n x) , \; x \in \mathbb{R}\) is the Fourier series of the function
\[f({x})=\left\{{\begin{array}{lc}
\frac{\pi}{2}\left({2-x+2\left\lfloor\frac{x-1}{2}\right\rfloor}\right)\,,&2k-1<{x}<2k+1\\
0\,,& x=2k+1
\end{array}}\right.,\,{k}\in\mathbb{Z}\]
which is discontinuous at \(x=2k+1\,,\;{k}\in\mathbb{Z}\). On the other hand the partial sums \[S_n(x):=\sum_{k=1}^{n} \frac{(-1)^{k+1}}{k}\sin (\pi k x) , \; x \in \mathbb{R}\] of the series are continuous functions. So, the series converges almost uniformly but not uniformly in \(\mathbb{R}\).