# Sum of series $\sin x + \sin 2x + \sin 3x + \cdots$

Please help me compute the sum of the series: $$\sin(x)+\sin(2x)+\sin(3x)+\cdots$$

-
Welcome to stack exchange. please consider adding what you've tried, as well as using TeX formatting next time. – nbubis Aug 24 '12 at 13:44
Why do you think it converges? – GEdgar Aug 24 '12 at 13:44
What happens if $x=0$? – S. Snape Aug 24 '12 at 13:56
At least, it converges when $\quad\large\displaystyle x = n\pi\,,\quad n \in {\mathbb Z}$ – Felix Marin Jun 29 '14 at 1:05

The series does not converge for all $x$. There are some $x$, for instance $x=0$ or $x=\pi$, for which the series converges to $0$, however if we consider $x=\frac \pi 2$ we find that our series is $1+0+-1+0+1+\cdots$ which does not converge.

If you think of the unit circle, imagine a line whose angle from the positive x-axis is the value of $x$. Then the x-coordinate of this point on the unit circle is the value of $\sin x$. Doubling the angle yields a point whose x-coordinate is $\sin 2x$. Tripling it yields a point whose x-coordinate is $\sin 3x$. Continuing this, you can see that the pattern will continue with varied positive and negative values, not approaching any particular limit unless the line representing an angle of $x$ was aligned with the positive or negative x-axis. (This is not rigorous, but can be made to be so.)

More technically, we have that $\lim_{n\to\infty} \sin nx =0$ iff $x=k\pi$ for some $k\in\Bbb Z$, so the series trivially converges to zero for such $x$ and diverges for all other $x$.

-

$$2\sin\frac{x}{2}\sin rx=\cos\frac{(2r-1)}{2}x-\cos\frac{(2r+1)}{2}x$$

Putting $r=1,2,\ldots,n-1,n$ we get,

$$2\sin\frac{x}{2}\sin x=\cos\frac{1}{2}x-\cos\frac{3}{2}x$$

$$2\sin\frac{x}{2}\sin 2x=\cos\frac{3}{2}x-\cos\frac{5}{2}x$$

$$\vdots$$

$$2\sin\frac{x}{2}\sin rx=\cos\frac{(2n-3)}{2}x-\cos\frac{(2n-1)}{2}x$$

$$2\sin\frac{x}{2}\sin nx=\cos\frac{(2n-1)}{2}x-\cos\frac{(2n+1)}{2}x$$

Adding we get, $2\sin\frac{x}{2}(\sin x+\sin 2x+...+\sin nx)=cos\frac{1}{2}x-\cos\frac{(2n+1)}{2}x=2\sin\frac{(n+1)x}{2}\sin\frac{nx}{2}$

So, $$\sin x+\sin 2x+\cdots+\sin nx=\frac{\sin\frac{(n+1)x}{2}\sin\frac{nx}{2}}{\sin\frac{x}{2}}$$

As $2\sin B\sin(A+2rB) = \cos(A+(2r-1)B) - \cos(A+(2r+1)B)$, we need to multiply with $2\sin B$ for $\sum_{r}\sin(A+2rB)$.

Here in this problem, $A=0, 2B=x$

Also the way Nerd-Herd has approached the problem,

$\sin rx$ = Imaginary part of $e^{irx}$

So, $\sum_{0 ≤ r ≤n}\sin rx=\sum_{0 ≤ r ≤n}$(Imaginary part of $e^{ix})$=Imaginary part of($\sum_{0 ≤ r ≤n}e^{ix}$)

Now, $$\sum_{0 ≤ r ≤n}e^{irx}= \frac{e^{(n+1)ix}-1}{e^{ix}-1}= e^{\frac{(n+1)ix}{2}}\frac{(e^{\frac{(n+1)ix}{2}}-e^{-\frac{(n+1)ix}{2}})}{e^{\frac{ix}{2}}(e^{\frac{ix}{2}}-e^{-\frac{ix}{2}})}$$

We know, $\sin y=\frac{e^{iy}-e^{-iy}}{2i}$ So,$e^{iy}-e^{-iy}=2i\sin y$

So,$\sum_{0 ≤ r ≤n}e^{irx}=e^{\frac{inx}{2}}\frac{2i\sin\frac{(n+1)x}{2}}{2i\sin\frac{x}{2}}=(\cos \frac{nx}{2}+i\sin \frac{nx}{2})\frac{\sin\frac{(n+1)x}{2}}{\sin\frac{x}{2}}$

So, imaginary part of $\sum_{0 ≤ r ≤n}e^{irx}=\sin \frac{nx}{2}\frac{\sin\frac{(n+1)x}{2}}{\sin\frac{x}{2}}$

$\sum_{0 ≤ r ≤n}\sin{rx}=\sin \frac{nx}{2}\frac{\sin\frac{(n+1)x}{2}}{\sin\frac{x}{2}}$

$\sum_{1 ≤ r ≤n}\sin{rx}=\sin \frac{nx}{2}\frac{\sin\frac{(n+1)x}{2}}{\sin\frac{x}{2}}$ as $\sin(0x)=0$

So, the either approach leads to a compact from provided $\sin\frac{x}{2}≠0$

If $\sin\frac{x}{2}=0$ i.e., $\frac{x}{2}=m\pi$ where m is some integer,

$=>x=2m\pi=>\sin sx= 0$ for any integer s.

By observation if $\sin x=0$ i.e., $x=m\pi$ where m is some integer, $\sin sx= 0$ for any integer s.

Then the sum is clearly 0 if $x=m\pi$.

-

Using complex number system:

$$\sin{x} = \text{Im} ( {e^{ix} )}$$ and since,

$$\text{Im} (z_2) + \text{Im} (z_2) = \text{Im} (z_1 + z_2)$$

Using this, you get a geometric progression. Which gives the result as

$$\text{Im} (\frac{e^{ix}}{1-e^{ix}})$$

-
Watch out: you just summed a geometric series whose ratio has modulus 1. This cannot be done. – Did Aug 24 '12 at 14:23
@sos440 If the series converges, it'll converge to the same value as given by $$\text{Im} (\frac{e^{ix}}{1-e^{ix}})$$ and if it doesn't converge; then it just doesn't. – hjpotter92 Aug 24 '12 at 14:27
@Nerd-Herd : your sum is wrong. Geometric progression is summable iif the ratio has modulus $<1$ which is not the case since $|e^{iz}|=1$. – vanna Aug 24 '12 at 14:30
@Nerd-Herd, that exactly means that it does not converge in ordinary sense. But there are many other generalized ways to assign a value to a formal series (indeed, ordinary summation sense is also a way of assigning a value to a formal series by defining its value as the limit of the partial sum), and some of them actually yield $$\sum_{n=1}^{\infty} \sin nx = \Im \left(\frac{e^{ix}}{1-e^{ix}}\right) = \frac{1}{2}\cot\frac{x}{2},$$including Cesaro summation sense and Abel summation sense. – Sangchul Lee Aug 24 '12 at 14:31
Nerd-Herd: If the series converges, I am Julius Caesar. – Did Aug 24 '12 at 15:15

The series converges but not in the usual sense if $x$ is real. If we assume that $x$ is real, then one might be tempted to write $$\sum_{n = 1}^\infty \sin nx \color{red}{=} \textbf{I}\bigg[\sum_{n = 1}^\infty e^{inx}\bigg] \color{red}{=} \textbf{I}\bigg[\frac{e^{ix}}{1 - e^{ix}}\bigg] \color{red}{=} \frac{\sin x}{2(\cos x - 1)}$$ and, similarly, $$\sum_{n = 1}^\infty \cos nx \color{red}{=} \textbf{R}\bigg[\sum_{n = 1}^\infty e^{inx}\bigg] \color{red}{=} \textbf{R}\bigg[\frac{e^{ix}}{1 - e^{ix}}\bigg] \color{red}{=} -\frac{1}{2}.$$

To find the real and imaginary parts multiply and divide $e^{ix}/(1 - e^{ix})$ by $1 - e^{-ix}$ and used Euler's formulae $e^{ix} = \cos x + i\sin x$ and $e^{ix} + e^{-ix} = 2\cos x$.

But $|e^{ix}| = 1$ if $x$ is real, so $\sum_{n = 1}^\infty e^{inx}$ diverges and the $\color{red}{\text{red}}$ equalities are metaphoric because we are assigning a value to each series. The value of a series is, in general, different from its sum.

However, the series $\sum_{n = 1}^\infty e^{inx}$ converges if $x$ is in the upper half-plane, that is, $\textbf{I}[x] > 0$, but it does not help because then $$\sum_{n = 1}^\infty \sin nx \ne \textbf{I}\bigg[\sum_{n = 1}^\infty e^{inx}\bigg].$$

-
Can the real part of a holomorphic function be constant? I don't think $\frac{-1}2$ is correct for the second sum. – Thomas Andrews Jun 28 '14 at 23:39

There is much more easier way to prove that the series does not converge when $x \ne k\pi$.

$\liminf\limits_{n\to\infty}\sin nx=-1$

The corresponding subsequence is $n_k=-\frac{\pi -4\pi k}{2x}$

$\limsup\limits_{n\to\infty}\sin nx=1$

The corresponding subsequence is $n_k=\frac{\pi + 4\pi k}{2x}$

Therefore there is no limit of $\sin nx$ and hence a necessary condition of convergence is violated.

-
You really mean a necessary condition, not the necessary condition. A necessary condition for a series to converge is that the limit of the terms converges to zero. – 6005 Jun 29 '14 at 4:46
Thank you for your comment. I've changed my answer. – Тимофей Ломоносов Jun 29 '14 at 5:01
Except this does not prove the liminf is -1 and the limsup is +1 since your choices of n_k are illegal most of the time (there is no reason that the RHS would be integers). – Did Jun 29 '14 at 6:55