Evaluate
$$\sum_{n=1}^\infty (-1)^{n+1} \left({1 - \cos(2n-1) \over 2n-1}\right)$$
Given
$f(x) = 1$ if $|x| < 1$ and $f(x) = 0$ if $1 \leqslant\ |x| \leqslant\ \pi$.
I calculated
$$Sf(x) = {1\over\pi} + {2\over\pi}\sum_{n=1}^\infty{{\sin(n)\over n} \cos(nx)}$$
And also calculated
$$\sum {{\sin(n) \over n} = {1\over 2}(\pi - 1)}$$
Which is asked to be used as an "inspiration" to solve the summation that I want.
But now I'm kinda lost on how to get to the sum in the title. Can someone at least point me in the right direction?
Edit:
From Yuriy S's suggestion I got
$${1\over\pi}(1 + 1\sum{\sin(1+x)+\sin(1-x)\over n})$$
But again, I don't know what $x$ to choose that will make me get to the desired sum.
Edit 2:
After A LOT of thinking about this series, I was finally able to answer it using the Fourier Series, simply by using the odd extension of the function, like so:
$f(x) = 1$ if $0 < x < 1$ and $f(x) = -1$ if $-1 < x < 0$.
This function will now give
$$Sf(x) = {2\over\pi}\sum_{n=1}^\infty{{1-\cos(n)\over n} \sin(nx)}$$
Now we have to force n to be an odd number, so we have
$$Sf(x) = {2\over\pi}\sum_{j=1}^\infty{{1-\cos(2j-1)\over 2j-1} \sin((2j-1)x)}$$
$$(n = 2j-1)$$
Choosing $x = {\pi\over 2}$ will give
$$Sf(x) = {2\over\pi}\sum_{j=1}^\infty{{1-\cos(2j-1)\over 2j-1} \ (-1)^{j+1}}$$
$Sf({\pi\over 2}) = 0$ will give
$$\sum_{j=1}^\infty{{1-\cos(2j-1)\over 2j-1} \ (-1)^{j+1}} = 0$$
Which is the same result as Felix Marin's solution to the problem.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align} &\bbox[10px,#ffd]{\ds{% \sum_{n = 1}^{\infty}\pars{-1}^{n + 1}\,{1 - \cos\pars{2n - 1} \over 2n - 1}}} = \Re\sum_{n = 1}^{\infty}\pars{-1}^{n + 1}\, {1 - \expo{\pars{2n - 1}\ic} \over 2n - 1} \\[5mm] = &\ \Re\bracks{\ic^{3}\sum_{n = 1}^{\infty}\ic^{2n - 1}\, {1 - \expo{\pars{2n - 1}\ic} \over 2n - 1}} = \Im\sum_{n = 1}^{\infty}\ic^{n}\, {1 - \expo{n\ic} \over n}\,{1 - \pars{-}^{n} \over 2} \\[5mm] = &\ \Im\sum_{n = 1}^{\infty} {\ic^{n} - \pars{\ic\expo{\ic}}^{n} \over n}\,{1 - \pars{-1}^{n} \over 2} = \Im\sum_{n = 1}^{\infty}{\ic^{n}\over n} - {1 \over 2}\Im\sum_{n = 1}^{\infty}{\pars{\ic\expo{\ic}}^{n} \over n} + {1 \over 2}\Im\sum_{n = 1}^{\infty}{\pars{-\ic\expo{\ic}}^{n} \over n} \\[5mm] = &\ -\,\Im\ln\pars{1 - \ic} + {1 \over 2}\,\Im\ln\pars{1 - \ic\expo{\ic}} - {1 \over 2}\,\Im\ln\pars{1 + \ic\expo{\ic}} \\[5mm] = &\ {\pi \over 4} + {1 \over 2}\,\Im\ln\pars{1 + \sin\pars{1} - \cos\pars{1}\ic} - {1 \over 2}\,\Im\ln\pars{1 - \sin\pars{1} + \cos\pars{1}\ic} \\[5mm] = &\ {\pi \over 4} + {1 \over 2}\,\arctan\pars{-\cos\pars{1} \over 1 + \sin\pars{1}} - {1 \over 2}\,\arctan\pars{\cos\pars{1} \over 1 - \sin\pars{1}} \\[5mm] = &\ {\pi \over 4} - {1 \over 2}\,\arctan\pars{\cos\pars{1} \over 1 + \sin\pars{1}} - {1 \over 2}\,\mrm{arccot}\pars{1 - \sin\pars{1} \over \cos\pars{1}} \\[5mm] = &\ {\pi \over 4}\ \underbrace{- {1 \over 2}\,\arctan\pars{\cos\pars{1} \over 1 + \sin\pars{1}} - {1 \over 2}\,\mrm{arccot}\pars{\cos\pars{1} \over 1 + \sin\pars{1}}} _{\ds{-\,{\pi \over 4}\quad\mrm{because}\quad {\cos\pars{1} \over 1 + \sin\pars{1}} > 0}}\ =\ \bbx{\large 0} \end{align}