Show that the Fourier series for $f(x) =x$ is given by $S_f(x) = 2(\sum_{k=1}^\infty \frac{\sin(2k-1)x}{2k-1}-\sum_{k=1}^\infty\frac{\sin(2kx)}{2k})$

Question

the interval of the Fourier series is $[-\pi,\pi]$, which didn't fit the title.

I have calculated the following Fourier coefficients: $a_0 = 0,a_k = 0, b_k = \frac{2*(-1)^{k+1}}{k}$ which means that the Fourier Series is given by: $$ S_f(x) = \sum_{k=1}^\infty \frac{2*(-1)^{k+1}}{k}*\sin(kx) $$ However, at this point, I have no idea how to prove that this is in fact equal to $S_f(x) = 2(\sum_{k=1}^\infty \frac{\sin(2k-1)x}{2k-1}-\sum_{k=1}^\infty\frac{\sin(2kx)}{2k})$.

Answer 1

The answer was basically given in the comments: $$\sum_{k=1}^{2N}(-1)^k a_k=\sum_{k=1}^N a_{2k}-\sum_{k=1}^N a_{2k-1}$$