Well, take the function f(x) = x, (-pi, pi), f(x+2(pi)n) = f(x), n E (here E is being used as "is an element of," not the letter) Z (all integers), (-infinity, infinity). Essentially, it is the line f(x) = x, from -pi to pi. At pi, it goes from pi to negative pi, creating a saw wave (think of it being |/|/|/|/|/ from negative infinity to infinity, where the bars are at pi*n).
Now, there's an interesting relationship between the coefficients of a fourier series (which is just using sine and cosine to represent a periodic function that isn't normally represented with sine and cosine, like our earlier saw wave) and the function itself:
sum(n=1, n -> infinity) (|cn(sum of the coefficients)|2) = int(-pi to pi)(f(x)2dx)/(2pi). In other words, the sum where n goes from 1 to infinity of the absolute value of the fourier coefficients squared is equal to the integral from -pi to pi of the function squared over 2pi. As for the coefficients:
bn = int(-pi to pi)(f(x)cos(nx)dx)/pi, n >= 0, which equals 0 (if f(x) = x, then it's zero, because the integral of an odd function from -a to a is zero)
an = int(pi to pi)(f(x)sin(nx)dx)/pi, n>=1, which equals (if you want, I'll solve the integral in a subsequent post) (2(-1)n+1)/n, n >= 1.
With that, we know that cn = an, since bn = 0. Therefore, the sums absolute value of (2(-1)n+1)/n squared from 1 to infinity should be equal to the integral from -pi to pi of x2 over 2 pi.
Well, the absolute value of an ends up just being 1/n (we pull out the two in front of the sum of the coefficients beforehand). Thus, the sum from one to infinity of 1/n2 is equal to the integral of x2 from -pi to pi, all over 4 pi. This, of course, ends up being pi2/6.