sum of :$\sum_{k=1}^\infty\frac{(-1)^k}{2k-1} \cos(2k-1)$

Question

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?

Answer 1

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 $\arctan(1)=\pi/4$.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#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 $\ds{\bf\color{black}{4.4.34}}$