Limit of summation of trigonometric series

Question

Let $S_n=\sum_{k=1}^n \sin(k\theta)$.

Prove that: $$\lim\limits_{n \to \infty} \sum_{k=1}^n \frac {S_k} n=\frac 1 2 (\cot( \theta/2))$$

Answer 1

If $e^{i\theta} \neq 1$ ($\theta \neq 0 \, (\textrm{mod } 2\pi)$), we have:

$$S_n=\sum_{k=1}^n \sin(k\theta)=\Im{\sum_{k=1}^n e^{ik\theta}}=\sin(\frac{n+1}{2}\theta)\frac{\sin(\dfrac{n}{2}\theta)}{\sin(\theta/2)}$$

Then $$\sum_{n=1}^N S_n=\dfrac{1}{\sin(\theta/2)}\sum_{n=1}^N \sin(\dfrac{n}{2}\theta)\sin(\frac{n+1}{2}\theta) = -\dfrac{1}{\sin(\theta/2)}\sum_{n=1}^N \dfrac{1}{2}(\cos(\dfrac{n+1}{2}\theta)-\cos(\theta/2)) $$

$$\sum_{n=1}^N S_n=\dfrac{N}{2\tan(\theta/2)}-\dfrac{1}{2\sin(\theta/2)}\sum_{n=1}^N\cos(\dfrac{n+1}{2}\theta)$$

And as previously you can compute

$$\sum_{n=1}^N\cos(\dfrac{n+1}{2}\theta)=\Re \sum_{n=1}^{N} e^{i\theta n/2}e^{i\theta/2}=\cos((N/2+1)\theta)\dfrac{\sin(N\theta/2)}{\sin(\theta/2)}$$

which is bounded by $1/\sin(\theta/2)$.

Finally we find that $$\dfrac{1}{N}\sum_{n=1}^N S_n \to \dfrac{1}{2}\cot(\theta/2) $$

And if $e^{i\theta}=1$, then $S_n=0$ and the sum converges to $0$ which is not $\cot(0)/2$...