Fourier Series for |x|: Convergence & Answers

[BearY]

Homework Statement

Find trigonometric Fourier series for $f(x)=|x|$, $x∈[−\pi, \pi]$, state points where $F(x)$ fail to converge to $f(x)$.

Homework Equations

$F(x) = \frac{a_0}{2}+\sum\limits_{n=1}^\infty a_ncos(\frac{n\pi x}{L})+b_nsin(\frac{n\pi x}{L})$
$a_n=\frac{1}{L}\int_{-L}^{L}f(x)cos(\frac{n\pi x}{L})dx$
$b_n=\frac{1}{L}\int_{-L}^{L}f(x)sin(\frac{n\pi x}{L})dx$

The Attempt at a Solution

$f(x)cos(nx)$ is an even function$a_n=\frac{2}{\pi}\int_{0}^{\pi}xcos(nx)dx$
$f(x)sin(nx)$ is an odd function, so $b_n = 0$
$$a_0 =\frac{2}{\pi}\int_{0}^{\pi}xdx = \pi$$
$$a_n=\frac{2}{\pi}(\frac{cos(n\pi)-1}{n^2})$$
$$F(x) = \frac{\pi}{2}+\sum\limits_{n=1}^\infty \frac{2}{\pi}(\frac{cos(n\pi)-1}{n^2})cos(nx)$$

My question is, the answer is
$$F(x) = \frac{\pi}{2}-\frac{4}{\pi}\sum\limits_{n=1}^\infty \frac{1}{(2n-1)^2}cos((2n-1)x)$$
I can see that the summation in my $F(x)$ has even terms equal to 0 and hence the $(2n-1)$. But I am not sure how to get there. Or maybe my answer is wrong.

For completion of answer$F(x)$ converge to $f(x)$ everywhere in the domain.

 

[I like Serena]

Hi BearY! :)

Split the summation in even n and odd n.
Note that $\cos(2k\pi)=1$, which cancels, while $\cos((2k-1)\pi)=-1$.