4

Solving a bigger problem about Fourier series I'm faced with this sum:

$$\sum_{k = 0}^{\infty} \frac{(-1)^k}{k} \sin(2k)$$

and I've no idea of how to approach this.

I've used Leibniz convergence criterium to verify that the sum should have a value, but I don't know how to calculate this value.

mathlove
  • 139,939
iveqy
  • 1,327

4 Answers4

3

Hint:

$e^{2ik}-e^{-2ik}=2i\sin 2k$

Vera
  • 3,469
3

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ Lets $\ds{\fermi\pars{x}\equiv \left\{\begin{array}{lcrcl} {\sin\pars{2x} \over x} & \mbox{if} & x & \not= & 0 \\[1mm] 2 & \mbox{if} & x & = & 0 \end{array}\right.}$

Note that $\ds{\fermi}$ is an even function of $\ds{x}$: $\ds{\fermi\pars{-x}=\fermi\pars{x}\,,\ \forall\ x\in{\mathbb R}}$.

We'll use the $\large\mbox{Abel-Plana Formula}$:

\begin{align} &\color{#66f}{\large\sum_{n = 1}^{\infty}\pars{-1}^{n}\,{\sin\pars{2n} \over n}} =-2 + \sum_{n = 0}^{\infty}\pars{-1}^{n}\fermi\pars{n} \\[5mm]&=-2 + \bracks{\half\,\fermi\pars{0} + \ic\ \underbrace{\int_{0}^{\infty}% {\fermi\pars{\ic t} - \fermi\pars{-\ic t} \over 2\sinh\pars{\pi t}}\,\dd t} _{\ds{=\ \color{#c00000}{\large 0}}}} \\[5mm]&=-2 + \bracks{\half\times 2 + \ic\times 0} = \color{#66f}{\Large -1} \end{align}

Felix Marin
  • 89,464
2

$$\sum_{k=1}^{+\infty}\frac{(-1)^k}{k}\sin(2k)=\Im\sum_{k=1}^{+\infty}\frac{(-1)^k}{k}e^{2ik}=-\Im\log(1+e^{2i})=\color{red}{-1}.$$

Jack D'Aurizio
  • 353,855
  • what is −ℑ? And how come you get -1 and Leucippus 1? – iveqy Oct 03 '14 at 15:36
  • @iveqy: $\Im$ is the imaginary part. My result and Leucippus' one differ because of the $k=0$ term, that is not included in my sum (since it looked not very well defined to me). – Jack D'Aurizio Oct 03 '14 at 15:53
2

Consider the series \begin{align} S = \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} \, \sin(2n). \end{align}

Method 1

Using the known Fourier series \begin{align} x = \frac{2 L}{\pi} \, \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \, \sin\left( \frac{n \pi x}{L} \right) \end{align} it can quickly be seen that for $L = \pi$ and $x = 2$ the series becomes \begin{align} - 1 = \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} \, \sin(2n). \end{align}

Method 2

Using $2i \sin(2n) = e^{2in} - e^{-2in}$ then the series is \begin{align} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} \, \sin(2n) &= \frac{1}{2i} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} \left( e^{2in} - e^{-2in} \right) \\ &= - \frac{1}{2i} \left( \ln(1 + e^{2i}) - \ln(1 + e^{-2i}) \right) \\ &= - \frac{1}{2i} \, \ln\left(\frac{1 + e^{2i}}{1 + e^{-2i}} \right) = - \frac{1}{2i} \, \ln\left(\frac{e^{i} \, \cos(1)}{e^{-i} \, \cos(1)} \right) \\ &= - \frac{1}{2i} \ln(e^{2i}) = -1. \end{align}

Method 3

As stated in the proposed problem the summation is given by \begin{align} S_{0} = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n} \, \sin(2n), \end{align} for which \begin{align} S_{0} &= \lim_{n \rightarrow 0} \left\{ \frac{\sin(2n)}{n} \right\} + \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} \, \sin(2n) \\ &= -1 + \lim_{n \rightarrow 0} \left\{ \frac{2 \cos(2n)}{1} \right\} \\ &= -1 + 2 = 1. \end{align}

Leucippus
  • 26,329
  • How come you get -1 for method 1 and 2 and 1 for method 3? – iveqy Oct 03 '14 at 15:41
  • @iveqy As stated in the proposed problem the summation index starts at zero. In method 1 and 2 the summation index starts at one. – Leucippus Oct 03 '14 at 15:44
  • Thanks for a good solution. Could you explain how you in method 2 do the step where $\sum$ disappears? – iveqy Oct 04 '14 at 06:36
  • @iveqy The series is transformed into a function, in this case the logarithm, by making use of the formula $\begin{align} \ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} , x^{n} \end{align}$. – Leucippus Oct 04 '14 at 15:02
  • You have an error in method 2, it's $\ln (1+e^{2i}) - \ln (1+e^{-2i})$, not $\ln (1+e^{2i}) - \ln (1-e^{-2i})$. – Daniel Fischer Oct 05 '14 at 14:13
  • @DanielFischer fixed the sign error typos. thanks for the edit help – Leucippus Oct 05 '14 at 14:43
  • is it really allowed to do the $\ln a - \ln b = \ln a/b$ in method 2 since it contains complex numbers? – iveqy Oct 05 '14 at 15:11
  • @iveqy In general, you have - for any specific branch of the logarithm - only $\ln a - \ln b = \ln \frac{a}{b} + 2k\pi i$ for some integer $k$. Here, one can see that for the principal branch of the logarithm (which is the one you get from the series), you have $k = 0$ [since $\lvert \operatorname{Im}\ln (1+e^{\pm 2i})\rvert < \frac{\pi}{2}$, as $\operatorname{Re} (1+e^{\pm 2i}) > 0$]. But yes, strictly, that identity requires a justification. – Daniel Fischer Oct 05 '14 at 15:37