Fourier series of $f$ in the interval $[-\pi, \pi]$ and evaluation of a series I have the function $$f(x) :=\begin{cases}1,& \frac{\pi}{2}< x < \pi
                           \\0, &-\frac{\pi}{2}< x < \frac{\pi}{2}
                          \\-1, &-\pi < x < -\frac{\pi}{2}\end{cases}$$
and I compute $a_0, a_n$ and $b_n$, $$a_0=a_n=0,\quad\text{and}\quad b_n = \frac{1}{\pi}\left(\frac{2}{k}\cos\big(\frac{\pi k}{2}\big) - \frac{2}{k}\cos(\pi k)\right)$$
Now I have a problem with the calculation of the next sum:
$$\sum_{n=1}^{\infty}\frac{(-1)^n}{(2n-1)}$$
I do not know how to compute this sum with $a_0,a_n$ and $b_n$. Thank you for your help!
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over 2n - 1} & =
\ic\sum_{n = 1}^{\infty}{\ic^{2n - 1} \over 2n - 1} =
\ic\sum_{n = 1}^{\infty}{\ic^{n} \over n} -
\ic\sum_{n = 1}^{\infty}{\ic^{2n} \over 2n} =
\ic\sum_{n = 1}^{\infty}{\ic^{n} \over n} -
{1 \over 2}\,\ic\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n}
\\[5mm] & =
-\ic\ln\pars{1 - \ic} - {1 \over 2}\,\ic\braces{-\ln\pars{1 - \bracks{-1}}} =
-\ic\bracks{{1 \over 2}\,\ln\pars{2} - {\pi \over 4}\,\ic} +
{1 \over 2}\,\ic\ln\pars{2}
\\[5mm] & = \bbx{-\,{\pi \over 4}}
\end{align}
