sum of :$\sum_{k=1}^\infty\frac{(-1)^k}{2k-1} \cos(2k-1)$ How can I find the sum of :$$\sum_{k=1}^\infty\frac{(-1)^k}{2k-1} \cos(2k-1)$$
I don't fully understand the parseval identity so I am asking if we can use it to find the sum, and if so, how I should use it.
Is there a Fourier series we know the convergence to a function that can help?
 A: Recall the Maclaurin series of arctangent, valid for $|z|\leq 1,$ $z\neq\pm i$:
$$
\arctan(z) = \sum_{k=0}^{\infty}\frac{(-1)^kz^{2k+1}}{2k+1}
$$
$$
-\arctan(z) = \sum_{k=1}^{\infty}\frac{(-1)^kz^{2k-1}}{2k-1}
$$Put $z=e^{i}$ and take the real part:
$$
\Re(-\arctan(e^i)) = \Re\left( \sum_{k=1}^{\infty}\frac{(-1)^k(e^{i})^{2k-1}}{2k-1}\right)=\sum_{k=1}^{\infty}\frac{(-1)^k\cos(2k-1)}{2k-1}
$$The LHS evaluates to $-1\cdot \pi/4$, since the argument (angle) is $1$ and $\arctan(1)=\pi/4$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}}}
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\begin{align}
&\bbox[10px,#ffe]{\sum_{k = 1}^{\infty}{\pars{-1}^{k} \over 2k -1}\cos(2k-1)} =
\ic\sum_{k = 1}^{\infty}{\ic^{2k - 1} \over 2k - 1}\cos(2k - 1)
\\[5mm] = &\
\ic\sum_{k = 1}^{\infty}{\ic^{k} \over k}\cos(k)\,
{1^{k} - \pars{-1}^{k} \over 2 } =
-\,\Im\sum_{k = 1}^{\infty}{\ic^{k} \over k}\cos(k) =
-\,\Im\sum_{k = 1}^{\infty}{\ic^{k} \over k}{\expo{\ic k} + \expo{-\ic k}
\over 2}
\\[5mm] = &\
-\,{1 \over 2}\,\Im\sum_{k = 1}^{\infty}{\pars{\ic\expo{\ic}}^{k} \over k} -
{1 \over 2}\,\Im\sum_{k = 1}^{\infty}{\pars{\ic\expo{-\ic}}^{k} \over k}
\\[5mm] = &\
{1 \over 2}\,\Im\ln\pars{1 - \ic\expo{\ic}} +
{1 \over 2}\,\Im\ln\pars{1 - \ic\expo{-\ic}}
\\[5mm] = &\
{1 \over 2}\,\Im\ln\pars{1 + \sin\pars{1} - \ic\cos\pars{1}} +
{1 \over 2}\,\Im\ln\pars{1 - \sin\pars{1} - \ic\cos\pars{1}}
\\[5mm] = &\
-\,{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] = &\
-\,{1 \over 2}\,\bracks{{\pi \over 2} - \arctan\pars{1 + \sin\pars{1} \over \cos\pars{1}}} -
{1 \over 2}\arctan\pars{\cos\pars{1} \over 1 - \sin\pars{1}}
\\[5mm] = &\
\color{red}{-\,{\pi \over 4}} +
{1 \over 2}\
\overbrace{\bracks{\arctan\pars{1 + \sin\pars{1} \over \cos\pars{1}} -
\arctan\pars{\cos\pars{1} \over 1 - \sin\pars{1}}}}
^{\ds{\ =\ \color{red}{0}}}\label{1}\tag{1}
\\[5mm] = &\ \bbx{-\,{\pi \over 4}}\
\approx -0.7854
\end{align}
The brackets in (\ref{1}) vanishes out because
$\ds{{1 + \sin\pars{1} \over \cos\pars{1}} - {\cos\pars{1} \over 1 - \sin\pars{1}} = \color{red}{0}}$.
See
A & S $\ds{\bf\color{black}{4.4.34}}$
