Disproving uniform convergence of $\sum_{n=1}^\infty \frac2{\pi} \frac1n \Big (1-\cos\Big(n\frac{\pi}2\Big) \Big)\sin(nx)$

Question

I have a function $f: [0,\pi] \rightarrow \mathbb R : f(x) = 1 $ for $x < \frac{\pi}2$ and otherwise $f(x) = 0$

I evaluated the fourier sine series of $f$ and ended up with the following series:

$$\sum_{n=1}^\infty \frac2{\pi} \frac1n \Big (1-\cos\Big(n\frac{\pi}2\Big) \Big)\sin(nx)$$

I plotted this series and it seems to me that this is not convergent uniformly. However, I am having a lot of troubles proving this. I tried to somehow simplify the sum by knowing that $(1-\cos(n\pi /2)) $ is either $0,1$ 0r $2$, but this did not get me very far for disproving the uniform convergence.

Any help on how to do this would be greatly appreciated!

Answer 1

HINT:

Write the series of interest as

$$\begin{align} \sum_{n=1}^\infty \frac{2\sin(nx)}{\pi n}(1-\cos(n\pi/2))&=\sum_{n=1}^\infty \frac{2\sin((2n-1)x)}{\pi (2n-1)}+\sum_{n=1}^\infty \frac{\sin(2nx)}{\pi n}(1-\cos(n\pi))\\\\ &=\sum_{n=1}^\infty \frac{2\sin((2n-1)x)}{\pi (2n-1)}+\sum_{n=1}^\infty \frac{2\sin(2(2n-1)x)}{\pi (2n-1)} \end{align}$$

and then apply Dirichlet's Test.

Answer 2

The sum converges pointwise to $$\tilde f(x) = \cases{ 1 & if $0\le x<\pi/2$,\cr 1/2 & if $x = \pi/2$,\cr 0 & if $\pi/2<x\le\pi$, }$$ But the uniform limit of continuous functions is continuous.