Evaluate $\sum_{n=1}^\infty (-1)^{n+1}({1 - \cos(2n-1) \over 2n-1})$ using Fourier Series 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.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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Besides the 'Fourier Technique', it's interesting to explore another ways !!!:

\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}
