6
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ I doubt Wolfram indicated that you have to do something to solve a certain problem $\endgroup$ Dec 21, 2017 at 22:23
  • $\begingroup$ Alright, I see your point. Re-worded. $\endgroup$
    – Alec
    Dec 21, 2017 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? $\endgroup$ Dec 21, 2017 at 22:33

5 Answers 5

9
$\begingroup$

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!

$\endgroup$
1
  • $\begingroup$ That's pretty neat! $\endgroup$
    – Alec
    Dec 21, 2017 at 23:49
6
$\begingroup$

Using

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

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 $\cos$

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Very nice, +1... $\endgroup$ Dec 22, 2017 at 11:14
  • $\begingroup$ @GabrielRomon Thank you. Much appreciative. $\endgroup$
    – Mark Viola
    Dec 22, 2017 at 14:33
1
$\begingroup$

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

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}

$\endgroup$
0
$\begingroup$

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 $$\begin{eqnarray*} \lim_{u\to 0^+}\sum_{n\geq 1}\frac{(-1)^n e^{-nu}\sin(n)}{n} &=&\sum_{n\geq 1}\frac{(-1)^n\sin(n)}{n}\\ &=& -\text{Arg}(1+e^i) = -\frac{1}{2}-\text{Arg}\left(2\cos\tfrac{1}{2}\right)=\color{red}{-\frac{1}{2}}.\end{eqnarray*}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.