Thanks SM, that's much closer. Should I conclude that there is nothing in Wolfram that would return a simplified/rationalised solution? In this case I was checking a result that I knew was at least close, but - for example - if I wanted a simple function out of a Fourier expansion rather than doing it analytically, I probably wouldn't know anything about it's likely form ....

Thanks, that specifies the function, but I want the output in the form: Sum (n=0 t inf) (sin x)/n ?

You are using Latex coding, which doesn't work in Mathematica.

Here is the way to define your function in Mathematica:

f[x_]:=Which[-Pi<=x<0,-1/2 (Pi+x), 0<x<=Pi,1/2 (Pi-x)]

You can plot this to see your sawtooth function. Plot[f[x], {x, -Pi, Pi}]

and you can find the Fourier transform

In[3]:= FourierTransform[f[x], x, k]

Out[3]= (E^(-I k \[Pi]) - E^(I k \[Pi]) + 2 I k \[Pi])/(2 k^2 Sqrt[
 2 \[Pi]])

Or if you want the complex Fourier series up to n=3

In[6]:= FourierSeries[f[x], x, 3]

Out[6]= 1/2 I E^(-I x) - 1/2 I E^(I x) + 1/4 I E^(-2 I x) - 
 1/4 I E^(2 I x) + 1/6 I E^(-3 I x) - 1/6 I E^(3 I x)