Derivatives and integrals of Fourier series

I've got the series and the derivatives and integrals. How can I show what they sum to?

Or can I simply say that the Integral sums to the integral of |x| for |x|< l because the terms decrease more quickly and so it converges.

If the derivative converges can I say likewise for x=-l,l,0 its undefined for 0<x<l it's 1 and -l<x<0 it's -1. Also how can I show it converges? as the terms decrease more slowly than for the original fourier series and so its not necessarily going to converge...

It is a fact that the Fourier series of the integral is the termwise integral of the Fourier series, and the Fourier series of the derivative is the termwise derivative of the Fourier series, assuming $\sum_{n=0}^{\infty} n (a_n + b_n)$ converges. All you need is uniform convergence of the series, and that condition turns out to be sufficient. More explicitly, you can use the Weierstrass M-test.

are those a_n and b_n of the original series?

can i quote $\sum_{n=0}^{\infty} n (a_n + b_n)$ as a standard result then as regards the term wise derivative?

I've got the series and the derivatives and integrals. How can I show what they sum to?

Or can I simply say that the Integral sums to the integral of |x| for |x|< l because the terms decrease more quickly and so it converges.

If the derivative converges can I say likewise for x=-l,l,0 its undefined for 0<x<l it's 1 and -l<x<0 it's -1. Also how can I show it converges? as the terms decrease more slowly than for the original fourier series and so its not necessarily going to converge...

More explicitly, you can use the Weierstrass M-test.

Not sure how the M-test works here, the obvious choice for M_n is O(1/n) and then $\sum M_n$ diverges.

Edit: in fact, I don't think the M-test *can* work here. If it does, then the sum converges uniformly, and so the result is continuous. But the derivative of |x| is not continuous...

FWIW, from wording of the question I don't think proof is required. There is actually a relevant theorem: given a continuous and piecewise-smooth function f, differentiating the Fourier series for f gives the correct Fourier series for f', but I think the proof is quite technical.

is |x| continuous and piecewise-smooth? and what are the criteria for it to be so?