Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
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 $x=0$? –  B. S. Aug 24 '12 at 13:56
add comment

3 Answers

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 $1+0+-1+0+1+\dots$ 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 $x$,

share|improve this answer
add comment

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

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

share|improve this answer
Watch out: you just summed a geometric series whose ratio has modulus 1. This cannot be done. –  Did Aug 24 '12 at 14:23
Actually, the given series does not converge in ordinary sense. your answer should be recognized as in a generalized summation sense, such as Cesaro summation or Abel summation. –  sos440 Aug 24 '12 at 14:24
@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. –  sos440 Aug 24 '12 at 14:31
Nerd-Herd: If the series converges, I am Julius Caesar. –  Did Aug 24 '12 at 15:15
show 1 more comment

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

Putting $r=1,2,...,n-1,n$ we get,

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

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


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

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

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

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 $x=m\pi$.

share|improve this answer
I'm wondering, are you aware that in MathJax $$..$$ produces a centered equation? –  user2468 Aug 25 '12 at 21:48
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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