# Calculating fourier series

I've a fourier series with a period = $2\pi$ that is even. f(t) = \begin{cases} 0 \text{, when: } 0<t<\pi-2 \\ \pi \text{, when: } \pi-2<t<\pi \end{cases} The functions trigonometric fourier series equals:

$2 + 2\sum_{k=1}^{\infty} (-1)^k\cos(kt)\cdot\frac{\sin(2k)}{k}$

What is the value of the series when $t = \pi-2$?

I tried to insert the value of t in the series and use Parevals formula and calculate the integral but I didn't manage to handle the integral. What should I do to calculate the sum of the series?

• You don't need to use the series. What do you know about the value of a fourier series at a point of discontinuity of the function? – Paul Oct 2 '14 at 14:21

I don't think you need to derive a 'bonus sum' using Parseval's formula here - note that the function satisfies Dirichlet's conditions on the open interval $(0,\pi)$ (prove this), and note that the function is discontinuous at $t= \pi -2$, and recall that the Fourier series converges to the average of the left and right limits at any point of discontinuity.