How to show that the Fourier's series of $f(x)=x$ uniformly converges?
After finding its coefficient, I got:
$$\sum\limits_{n=1}^{+\infty}\frac{2(-1)^{n+1}}{n}\sin(nx)$$
I showed the pointwise convergence.
$\frac{u_{n+1}}{u_n}=-\frac{n}{n+1}\frac{\sin((n+1)x)}{\sin(nx)}$
- And I'm not even sure that's always negative. But I assume that it show its decreasing.
- $\lim\limits_{n\longrightarrow+\infty}\frac{2(-1)^{n+1}}{n}\sin(nx)=0$
- And as far as $u_n$ is an alternated series
By Leibniz's rule $\sum u_n$ converges pointwise.
I want to study the uniform convergence, but I don't know how to manage it... can you give a method?
The uniform limit of countinuous functions is a countinuous function and in your case the (pointwise) limit is...