I'm trying to evaluate the following sum:

$$\displaystyle\sum\limits_{n=1}^{\infty}\frac{(-1)^n \sin(n)}{n}$$

But I'm having trouble getting anywhere. Wolphram Alpha indicated some stragety that uses complex logarithms in order to find the answer, which seems to be $-0.5$.

Anyone know a good strategy here?

  • $\begingroup$ I doubt Wolfram indicated that you have to do something to solve a certain problem $\endgroup$ – mathworker21 Dec 21 '17 at 22:23
  • $\begingroup$ Alright, I see your point. Re-worded. $\endgroup$ – Alec Dec 21 '17 at 22:26
  • $\begingroup$ Perhaps this is a Fourier series of some function at a particular point? A common trick with power series is to recognize them as Taylor series of an analytic function at a particular point, so you can evaluate just the function. Does something similar work here with maybe $x = 1/2\pi$ or something? $\endgroup$ – Alfred Yerger Dec 21 '17 at 22:33

Running with my initial comment, it turns out there is a way to make this strategy work.

Consider the function $s(x) = x/\pi$ when $x \in (- \pi, \pi)$, and extend this function to be periodic of period $2 \pi$ on the whole real line. This function is piecewise smooth, and so has a Fourier series, which is calculated in this wikipedia link. The resulting series is $$s(x) = \frac{2}{\pi}\sum_{n=1}^\infty \frac{(-1)^{n+1}\sin(nx)}{n} = -\frac{2}{\pi}\sum_{n=1}^\infty \frac{(-1)^n\sin(nx)}{n}$$

Which is a number times your sum. Denoting this sum as $S(x)$, and evaluating $s(x)$ at $1$, we get $\frac{1}{\pi}$, so $$s(1) = -\frac{2}{\pi}S(1) = \frac{1}{\pi}$$

This implies the answer Wolfram Alpha gave!

  • $\begingroup$ That's pretty neat! $\endgroup$ – Alec Dec 21 '17 at 23:49


$$\sin(x) = \Im( \exp(i x))\tag{1}$$

and for $|z|\le 1, z\ne 1$

$$\sum_{n=1}^{\infty}(-1)^n z^n/n=-\log(1+z)\tag{2}$$

we can write

$$ \begin{align} &s = \sum_{n=1}^{\infty} (-1)^n \sin(n) / n \\ &= \Im\sum_{n=1}^{\infty} (-1)^n \exp(i\; n) / n \\ &= - \Im \log (1+\exp(i))\\ & = - \frac{1} {2i}\left( \log (1+\exp(i))- \log (1+\exp(-i))\right)\\ &= - \frac{1} {2i} \log \left(\frac{ 1+\exp(i)}{1+\exp(-i)}\right)\\ &= - \frac{1} {2i} \log \left(\exp(i)\frac{ 1+\exp(i)}{\exp(i)+1}\right)\\ &= - \frac{1} {2i} \log \left(\exp(i)\right)\\ &= - \frac{1} {2i} i = -\frac{1}{2} \end{align} $$

Additional question:

Calculate the explicit expresion for the sum with $\sin$ replaced by $\cos$


I thought it might be instructive to present an approach that relies on straightforward, first year calculus methodologies. To that end we now proceed.

Note that we can write

$$\begin{align} \sum_{n=1}^N \frac{(-1)^n\sin(n)}{n}&=\int_0^1 \sum_{n=1}^N (-1)^n\cos(nx)\,dx\\\\ &=\int_0^1 \frac12 \left((-1)^N\cos((N+1)x)+(-1)^N\tan(x/2) \sin((N+1)x)-1\right) \,dx\tag1\\\\ &=-\frac12 +\frac12(-1)^N\,\frac{\sin(N+1)}{N+1}+\frac12(-1)^N\int_0^1 \tan(x/2)\sin((N+1)x)\,dx\tag2\\\\ &=-\frac12 +\frac12(-1)^N\,\frac{\sin(N+1)}{N+1}\\\\&+\frac12(-1)^N\left(-\frac{\tan(1/2)\cos((N+1))}{N+1}+\frac{1}{2(N+1)}\int_0^1 \sec^2(x/2)\cos((N+1)x)\,dx\right)\tag3 \end{align}$$

In arriving at $(1)$ we used the fact that $\cos(nx)=\text{Re}(e^{inx})$, then summed the geometric series $\sum_{n=1}^N (-e^{ix})^n$ and took the real part.

In going from $(2)$ to $(3)$ we integrated by parts with $u=\tan(x/2)$ and $v=-\frac{\cos((N+1)x)}{N+1}$.

Taking the limit as $N\to \infty$ in $(3)$ reveals

$$\begin{align} \sum_{n=1}^\infty \frac{(-1)^n\sin(n)}{n}&=\lim_{N\to \infty}\sum_{n=1}^N \frac{(-1)^n\sin(n)}{n}\\\\ &=-\frac12 \end{align}$$

And we are done!

Tools Used: Euler's formula, summing a geometric series, and integration by parts.

  • $\begingroup$ Very nice, +1... $\endgroup$ – Gabriel Romon Dec 22 '17 at 11:14
  • $\begingroup$ @GabrielRomon Thank you. Much appreciative. $\endgroup$ – Mark Viola Dec 22 '17 at 14:33

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

With the Abel-Plana Formula:

\begin{align} \sum_{n = 1}^{\infty}{\pars{-1}^{n}\sin\pars{n} \over n} & = -1 + \sum_{n = 0}^{\infty}\pars{-1}^{n}\,\mrm{sinc}\pars{n} = -1 + {1 \over 2}\,\mrm{sinc}\pars{0} = -1 + {1 \over 2} = \bbx{-\,{1 \over 2}} \end{align}

$\ds{\LARGE b)}$ \begin{align} \sum_{n = 1}^{\infty}{\pars{-1}^{n}\sin\pars{n} \over n} & = \Im\sum_{n = 1}^{\infty}{\pars{-1}^{n}\expo{\ic n} \over n} = \Im\sum_{n = 1}^{\infty}{\pars{-\expo{\ic}}^{n} \over n} = -\,\Im\ln\pars{1 + \expo{\ic}} \\[5mm] & = -\,\Im\ln\pars{1 + \cos\pars{1} + \sin\pars{1}\,\ic} = -\arctan\pars{\sin\pars{1} \over 1 + \cos\pars{1}} \\[5mm] & = -\arctan\pars{2\sin\pars{1/2}\cos\pars{1/2} \over 2\cos^{2}\pars{1/2}} = -\arctan\pars{\tan\pars{1 \over 2}} = \bbx{-\,{1 \over 2}} \end{align}


For any $|z|<1$ we have $\sum_{n\geq 1}\frac{(-1)^n z^n}{n}=-\log(1+z)$, hence by picking $z=e^{-u}e^{i}$ with $u>0$ we get $$ \sum_{n\geq 1}\frac{(-1)^n e^{-nu}\sin(n)}{n} = -\text{Im}\log(1+e^{-u}e^{i})=-\text{Arg}(e^u+e^{i}). $$ The sequences $\{\sin n\}_{n\geq 1}$ and $\{(-1)^n \sin n\}_{n\geq 1}$ have bounded partial sums, hence by summation by parts we are allowed to state that $$\begin{eqnarray*} \lim_{u\to 0^+}\sum_{n\geq 1}\frac{(-1)^n e^{-nu}\sin(n)}{n} &=&\sum_{n\geq 1}\frac{(-1)^n\sin(n)}{n}\\ &=& -\text{Arg}(1+e^i) = -\frac{1}{2}-\text{Arg}\left(2\cos\tfrac{1}{2}\right)=\color{red}{-\frac{1}{2}}.\end{eqnarray*}$$


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