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"Summing" the series $\sin(x)-\dfrac{1}{2}\sin(2x)+\dfrac{1}{3}\sin(3x)-\dfrac{1}{4}\sin(4x)+...$

Pose $$S=\sin(x)-\dfrac{1}{2}\sin(2x)+\dfrac{1}{3}\sin(3x)-\dfrac{1}{4}\sin(4x)+...$$ $$C=\cos(x)-\dfrac{1}{2}\cos(2x)+\dfrac{1}{3}\cos(3x)-\dfrac{1}{4}\cos(4x)+...$$

$$C+iS = e^{ix}-\dfrac{1}{2}(e^{ix})^2+\dfrac{1}{3}(e^{ix})^3-\dfrac{1}{4}(e^{ix})^4+...$$

Let $t$ = $e^ix$

Then we have a series of $t-\dfrac{t}{2}+\dfrac{t}{3}-\dfrac{t}{4}+...=\log(1+t)$

Which is $\log(1+e^{ix})=\log(1+\cos(x)+i\sin(x))$, use the formula $\log(A+iB)=\dfrac{1}{2}\log(A^2+B^2)+\arctan\left(\dfrac{B}{A}\right)$

$\log([1+\cos(x)]^2+i\sin(x))=\dfrac{1}{2}\log(\log([1+\cos(x)]^2+i\sin^2(x))+\arctan\left(\dfrac{i\sin(x)}{1+\cos(x)}\right)$. Since we are interested only in the imaginary part, we have the sum for $S$ is:

$$\arctan\left(\dfrac{i\sin(x)}{1+\cos(x)}\right)$$

I don't know what to do next.

In his paper "Subsidium Calculi Sinuum", Euler wrote the series has the "sum"

$$\sin(x)-\dfrac{1}{2}\sin(2x)+\dfrac{1}{3}\sin(3x)-\dfrac{1}{4}\sin(4x)+...=\dfrac{x}{2}$$

We obtained this by multiplying $dx$ integrating (his words is Illa autem series per $dx$ multiplicata et integrata dat:"

$$\cos(x)-4\cos(2x)+9\cos(3x)-16\cos(4x)...=0$$

I don't know how he obtains $\dfrac{x}{2}$, since the left hand side is $0$, how can it be $x/2$. Returning to my own "sum", how can I obtain the $1/2$

This series is quite important because it appears in Fourier "Analytical Theory of Heat".

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    $\begingroup$ note that $S$ is the imaginary part so it is $\arctan\left(\dfrac{\sin(x)}{1+\cos(x)}\right)$ (without $i$) and that is trivially $x/2$ for $ x \ne \pm \pi$ (modulo $2\pi$) by easy trigonometric manipulations; for $x = \pm \pi$ the sum is obviously zero; looking at the Fourier series of the odd function $x/2$ on $[-\pi, \pi]$ extended by periodicty to $\mathbb R$ also gives the result by the usual theorems about convergence of the Fourier series of a piecewise smooth periodic function $\endgroup$
    – Conrad
    Feb 28, 2021 at 2:00

2 Answers 2

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You are almost there! We have:

$$\ln(1 + \cos(x) + i\sin(x))= \ln(A + iB) = \frac{1}{2}\ln(A^{2}+B^{2})+i\arctan\bigg(\frac{B}{A}\bigg)$$

$$=\frac{1}{2}\ln\big((1 + \cos(x))^{2} + \sin^{2}(x)\big)+i\arctan\bigg(\frac{\sin(x)}{1 + \cos(x)}\bigg)$$

Then, the imaginary part is:

$$\arctan\bigg(\frac{\sin(x)}{1 + \cos(x)}\bigg)$$

Using the identity $\displaystyle\tan\bigg(\frac{x}{2}\bigg) = \frac{\sin(x)}{1 + \cos(x)}$:

$$ = \arctan\bigg(\tan\bigg(\frac{x}{2}\bigg)\bigg) = \boxed{\frac{x}{2}}$$

Of course, with this approach you will run into the usual domain problems with inverse trig functions and complex logarithms.

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Note

$$S’(x) = \cos x - \cos2x +\cos 3x -\cos 4x+ \cdots\\ = Re(e^{i x} - e^{i 2x} + e^{i 3x} -e^{i 4x} + \cdots) = Re \frac 1{1+ e^{-i x}}= \frac 12 $$

Thus

$$S(x) = \int_0^x S’(t) dt = \frac12x$$

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  • $\begingroup$ this is not quite rigurous as $\cos x - \cos2x +\cos 3x -\cos 4x+..=1/2$ is true only in a summation sense (Abel will do) but not in a convergence sense as obviously LHS is divergent for all $x$; one can argue from there but it requires a bit of theory about Fourier series of measures for example to make sense of LHS and to ensure that integrating term by term is correct (it is of course but needs proof) $\endgroup$
    – Conrad
    Feb 28, 2021 at 2:03
  • $\begingroup$ @Conrad, yes you are right. This is essentially the method Euler used to arrive at the series in my question. $\endgroup$
    – James Warthington
    Feb 28, 2021 at 16:20

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