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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?

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The uniform limit of countinuous functions is a countinuous function and in your case the (pointwise) limit is...

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