The limit of $2\frac{\sin\left(\frac{M\theta_{k}}{2}\right)\cos\left(\frac{1}{2}(1+M)\theta_{k}\right)}{\sin\left(\frac{\theta_{k}}{2}\right)} $

Question

I have

$$\lambda_{k,N}=2\dfrac{\sin\Big(\dfrac{M\theta_{k}}{2}\Big) \cos\Big(\dfrac{1}{2}(1+M)\theta_{k}\Big)}{\sin\Big(\dfrac{\theta_{k}}{2}\Big)} $$

where $\theta_{k} \in [0, 2\pi]$, I can then let $\theta_{k}=\dfrac{2\pi k}{N}$ for $k=0...N$,

I would like to find the limit of $\lambda_{k,N}$ when $N \to +\infty$ knowing that $M=M(N)$ and $\dfrac{M}{N}=0$ when $N \to +\infty$.

it would have been more interesting if I can compute the limit of $\lambda_{k,N}$ for any $k$ but it's not computable for some values of $\theta_{k}$.

Is it possible to find a closed form expression for the limit ?

Answer 1

Using the identity \begin{equation} \sin A\cos B=\frac{1}{2}[\sin(A+B)+\sin{(A-B)}] \end{equation} we get \begin{eqnarray} \lambda_{kN}&=&\frac{\sin\left[\left(M+\frac{1}{2}\right)\theta_k\right]-\sin\left(\frac{\theta_k}{2}\right)}{2\sin\left(\frac{\theta_k}{2}\right)}\\ &=&\frac{\sin\left((2M+1)\frac{\pi k}{N}\right)-\sin\left(\frac{\pi k}{N}\right)}{2\sin\left(\frac{\pi k}{N}\right)}\\ &=&\frac{\sin\left((2M+1)\frac{\pi k}{N}\right)}{2\sin\left(\frac{\pi k}{N}\right)}-\frac{1}{2} \end{eqnarray} So for any fixed $k$ \begin{equation} \lim_{N\to\infty}\lambda_{kN}=\lim_{n\to\infty}M(N)+\frac{1}{2}-\frac{1}{2}=\lim_{n\to\infty}M(N) \end{equation} provided the limit exists.