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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?

user376343
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Shane
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    Parseval identity is a statement about norm, and is a sort of pythagorean theorem with infinitly many squared coefficients. As they are squared, all terms will be non-negative, so your series cannot be deduced from Parseval identity I guess – Didier May 27 '20 at 14:32
  • yes that's right. do i need to look for a function that has a Fourier series? – Shane May 27 '20 at 14:40
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    Isn't $x$ missing in the cosine? – user376343 May 27 '20 at 14:43
  • One easy way to do it is to expand $1$ in sine series on $[0, \pi]$ by extending it to be an odd function on $[-\pi, 0]$ and then take $x=\pi/2-1$ as $\sin ((2k+1)(\pi/2-1))=(-1)^k\cos(2k+1))$; – Conrad May 27 '20 at 14:44
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    @user376343 no it's not missing – Shane May 27 '20 at 14:45
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    One possibility is to consider it as the real part of a power series in the complex domain – Damien May 27 '20 at 14:46
  • Mathematica evaluates the sum as $\sum\limits_{k=1}^\infty\frac{(-1)^k}{2 k-1} \cos (2 k-1)=-\frac{\pi }{4}$ which provides the result but not much insight as to the derivation. – Steven Clark May 27 '20 at 14:51
  • i've tried it too and saw that -pi/4 is the sum but how i get it? im trying to use the Fourier series of |x| now – Shane May 27 '20 at 15:00

2 Answers2

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

Integrand
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  • thank you, is there the possibility of using the series fourier of |x| to get to the same result? – Shane May 27 '20 at 15:09
  • @Shane better to use the fourier series of the rectangular wave – Integrand May 27 '20 at 15:21
  • @Integrand Sorry, I can not figure out how you obtain $$\arctan(e^i) = \arctan \left[ \cos(1) + i \sin(1) \right]$$ How to determine it? – BowPark May 27 '20 at 20:09
  • @BowPark splitting into real and imaginary parts, we have $$ \arctan(e^i)=\frac{1}{2} \arctan\left(\frac{\cos (1)}{1-\sin (1)}\right)+\frac{1}{2} \arctan\left(\frac{\cos (1)}{1+\sin (1)}\right)+i \left(\frac{1}{4} \log \left(2-2 \sin (1)\right)-\frac{1}{4} \log \left(2+2 \sin (1)\right)\right) $$Take the real part and use the arctan angle-addition formula. – Integrand May 27 '20 at 20:25
  • @BowPark A nicer way: if $\theta$ is real, $\Re\left(\arctan(e^{i\theta}) \right)= \pi/4\cdot \text{sign}(\cos(\theta))$. – Integrand May 27 '20 at 20:34
  • @Integrand I am not able to follow your steps. Sorry. I posted a dedicated question. – BowPark May 28 '20 at 09:27
  • How do you show that series converges for $|z|=1$ except when $z=\pm i$. I understand that $\arctan z$ has singularities at $z=\pm i$ so that radius of convergence is $1$ thus series converges if $|z|<1$. The case $|z|=1$ requires more analysis. – Paramanand Singh Jun 19 '20 at 09:04
  • @Paramanand Singh Dirichlet's Theorem – Integrand Jun 19 '20 at 13:23
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    Oh yes, I infact used the same Dirichlet test in a question linked with the current question being discussed here. – Paramanand Singh Jun 19 '20 at 13:50
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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 $\ds{\bf\color{black}{4.4.34}}$

Felix Marin
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