# How to show that the limit $\lim_{n \to \infty}\sin \frac{n \theta}{2}\sin \frac{(n+1) \theta}{2}$ doesn't exist?

I've been trying to show that $\sum_n \sin n\theta$ diverges (for most thetas at least), and I've come up with this expression for the partial sum (up to a multiplicative constant), and now I want to show that its limit doesn't exist:

$$\lim_{n \to \infty}\sin \frac{n \theta}{2}\sin \frac{(n+1) \theta}{2}$$

But I don't know how to proceed with this. Both terms are divergent, but that doesn't mean their product necessarily diverges (though in this case it sure seems so). Is there a straightforward way to show this limit doesn't exist?

EDIT: I want to clarify that while I did originally set out to show the divergence of a series, that's not the aim of this question, which is how to rigorously show a limit doesn't exist. I can show that the limit doesn't equal $0$, but I want to learn how to show that it can't equal any other number as well.

• You could use the following : If $\sum_{n} a_{n}$ converges, then $a_{n} \mathop{\longrightarrow} \limits_{n \to +\infty} 0$. – pitchounet May 2 '14 at 14:29

Note that $\sin\alpha\sin\beta=\frac12(\cos(\alpha-\beta)-\cos(\alpha+\beta))$, hence $$\sin\frac{n\theta}2\sin\frac{(n+1)\theta}2=\frac12\left(\cos\frac\theta2-\cos\bigl((n+\tfrac12)\theta\bigr)\right)$$ so unless $\theta$ is special ...
For a series $\sum a_n$ to converge it is nessecary that $\lim\limits_{n\to\infty }a_n=0$. Since this is not the case for $\sin(n\theta)$ (unless you choose $\theta=k\pi$) you have divergence here.
• @DepeHb So bassically you want a proof of the statement $\sum a_n$ convergent $\Rightarrow a_n = 0$. Is that correct? – gebruiker May 2 '14 at 14:45
• @DepeHb If you like you can prove it using the formal definition of a limit, but the way I did this, I used that I can write the $(k+1)^\text{st}$ term as the sum of the $k^\text{th}$ term and $\sin(k\theta)$. This is something you also had to do when you came up with the expression for the partial sum. Moreover it is, as I was implying before, prety much the proof of the forementioned theorem... – gebruiker May 2 '14 at 15:16