\(\sum_{n=1}^{+\infty}({-1})^{n+1}\frac{\sin{n}}{n}\)

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Grigorios Kostakos
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\(\sum_{n=1}^{+\infty}({-1})^{n+1}\frac{\sin{n}}{n}\)

#1

Post by Grigorios Kostakos »

Find \(\displaystyle\mathop{\sum}\limits_{n=1}^{+\infty}({-1})^{n+1}\frac{\sin{n}}{n}\,.\)
Grigorios Kostakos
ZardoZ
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Re: \(\sum_{n=1}^{+\infty}({-1})^{n+1}\frac{\sin{n}}{n}\)

#2

Post by ZardoZ »

Consider the sawtooth wave \(f(x)=x,\;\;x\in(-\pi,\pi)\) and \(f(x+2n\pi)=f(x)\) for \(x\in \mathbb{R}\) and \(n\in\mathbb{Z}\).
sawtoothwave.gif
sawtoothwave.gif (5.19 KiB) Viewed 3900 times
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Recall the fourier series formula \(\displaystyle f(x)=\frac{a_{0}}{2}+\sum_{k=1}^{N}\left[a_{k}\cos(kx)+b_{k}\sin(kx)\right]\), with \(\displaystyle a_{n}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)\;\mathbb{d}x\) for \(n\geq 0\) and \(\displaystyle b_{n}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)\;\mathbb{d}x\) for \(n\geq 1\). We can easily compute \(\displaystyle a_{n}=\frac{1}{\pi}\int_{-\pi}^{\pi}x\cos(nx)\;\mathbb{d}x=0\) and \(\displaystyle b_{n}=\frac{1}{\pi}\int_{-\pi}^{\pi}x\sin(nx)\;\mathbb{d}x=\frac{2 (\sin (n\pi )-\pi n \cos (n\pi ))}{ n^2\pi}\) which is equal to \(\displaystyle 2\frac{(-1)^{n}}{n}\) since \(n\in\mathbb{N}\). So the fourier series representation for \(f\) is \(\displaystyle f(x)=2\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin(nx)\) and \(\displaystyle f(1)=2\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\sin(n)}{n}\Rightarrow \sum_{n=1}^{\infty}\frac{(-1)^{n+1}\sin(n)}{n}=\frac{1}{2}\).
Vangelis Mouroukos
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Re: \(\sum_{n=1}^{+\infty}({-1})^{n+1}\frac{\sin{n}}{n}\)

#3

Post by Vangelis Mouroukos »

Another approach uses the power series expansion of the complex logarithm: For all \(z \in \Bbb{C}\) with \(|z| < 1\), we have
\[ \boxed{\displaystyle \log(1+z) = \sum_{n=1}^{\infty} (-1)^{n+1}\frac{z^n}{n}}\quad {\color {red} {(1)} }.\]
Substitute \(z = e^{i\theta} \), with \(\theta \in (-\pi, \pi) \), in \( \color {red} {(1)} \), to obtain
\[\log \left( {1 + {e^{i\theta }}} \right) = \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n + 1}}\dfrac{{{e^{in\theta }}}}{n}} \Longrightarrow \]
\[\Longrightarrow \log \left| {1 + {e^{i\theta }}} \right| + i{\rm{Arg}}\left( {1 + {e^{i\theta }}} \right) = \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n + 1}}\dfrac{{\cos n\theta }}{n}} + i\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n + 1}}\dfrac{{\sin n\theta }}{n}} \quad\color {red} {(2)} .\]
Separating real and imaginary parts in \( \color {red} {(2)} \), we obtain
\[\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n + 1}}\frac{{\cos n\theta }}{n}} = \log \left| {1 + {e^{i\theta }}} \right|\]
and
\[\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n + 1}}\frac{{\sin n\theta }}{n} = } {\rm{Arg}}\left( {1 + {e^{i\theta }}} \right).\]
Note that, for \(\theta \in \left( { - \pi ,\pi } \right)\), we have \(2\cos \dfrac{\theta }{2} > 0\) and
\[1 + {e^{i\theta }} = {e^{i\frac{\theta }{2}}}\left( {{e^{i\frac{\theta }{2}}} + {e^{ - i\frac{\theta }{2}}}} \right) = 2\cos \frac{\theta }{2}{e^{i\frac{\theta }{2}}}.\]
Hence, \(\left| {1 + {e^{i\theta }}} \right| = 2\cos \dfrac{\theta }{2}\) and \({\rm{Arg}}\left( {1 + {e^{i\theta }}} \right) = \dfrac{\theta }{2}.\)
It follows that
\[\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n + 1}}} \frac{{\cos n\theta }}{n} = \log \left( {2\cos \frac{\theta }{2}} \right) \quad{\color {red} {(3)}}\]
and
\[\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n + 1}}} \frac{{\sin n\theta }}{n} = \frac{\theta }{2} \quad{\color {red} {(4)}}.\]
In particular, for \({\theta = 1}\) we obtain the desired result:
\[\boxed{\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n + 1}}} \dfrac{{\sin n}}{n} = \dfrac{1}{2}}.\]
It follows from Abel's Test that the convergence in \(\color {red} {(3)}\) and \(\color {red} {(4)}\) is uniform on each interval of the form \(\left[ { - \pi + \delta ,\pi - \delta } \right]\), where \(0 < \delta < \pi \).
One can similarly obtain that, for \(\theta \in \left( {0,2\pi } \right)\),
\[\sum\limits_{n = 1}^\infty {\dfrac{{\cos n\theta }}{n}} = - \log \left( {2\sin \dfrac{\theta }{2}} \right)\]
and
\[\sum\limits_{n = 1}^\infty {\dfrac{{\sin n\theta }}{n}} = \dfrac{{\pi - \theta }}{2},\]
the convergence being uniform on each interval of the form \(\left[ {\delta ,2\pi - \delta } \right]\), where \(0 < \delta < \pi \).
ZardoZ
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Re: \(\sum_{n=1}^{+\infty}({-1})^{n+1}\frac{\sin{n}}{n}\)

#4

Post by ZardoZ »

A brief solution with the help of complex analysis techniques would be by defining the function \(\displaystyle f(z)=\frac{\pi \csc(\pi z)\sin(z)}{n}\), then

\[\sum_{n=-\infty,\,n\neq 0}^{\infty}\frac{(-1)^{n}\sin(n)}{n}=-\textrm{Residue}\left(f(z);z=0\right)=-1\Rightarrow\sum_{n=-\infty,\,n\neq 0}^{\infty}\frac{(-1)^{n+1}\sin(n)}{n}=1 \]

and so \(\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n+1}\sin(n)}{n}=\frac{1}{2}\).
akotronis
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Re: \(\sum_{n=1}^{+\infty}({-1})^{n+1}\frac{\sin{n}}{n}\)

#5

Post by akotronis »

A nice relevant problem, proposed by Ovidiu Furdui, that is still open for submitting solutions is H722 here
http://www.fq.math.ca/Problems/2012AugustAdvanced.pdf
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