Maximizing the sum $\sum_{n=1}^m \sin n$

Question

Consider the sum : $$\displaystyle \sum_{n=1}^m \sin n$$ For which value of $m,$ we will obtain the maximum sum?

Here's my approach : $\displaystyle \sum_{n=1}^m \sin n=\dfrac{\sin 1}{4 \sin^2 \dfrac{1}{2}} - \dfrac{2\cos \left( m+\dfrac{1}{2} \right)}{4 \sin \left( \dfrac{1}{2}\right)}$

If we can minimize $2 \cos \left(m+\dfrac{1}{2} \right)$ then this will result in maximizing the sum.

But the problem is I can't quite figure out what the minimum value of $\cos \left(m+\dfrac{1}{2}\right)$ is.

Answer 1

Hint: $\cos(m+1/2)$ is close to $1$ if $m+1/2$ is close to a multiple of $2\pi$.

Can you see why the values of $m+1/2$ get close to $2\pi$ using rational approximations?

Answer 2

There is no value $m$ which solves your conditions.

For a maximum we need $\,\displaystyle m:=\pi(2k+1)-\frac{1}{2}\in\mathbb{N}\,$ which is not possible for any $\,k\in\mathbb{Z}\,$ .

The supremum for the sum is $\,\displaystyle \dfrac{\sin 1}{4 \sin^2 \dfrac{1}{2}} + \dfrac{1}{2 \sin \left( \dfrac{1}{2}\right)}\,$ .

Answer 3

$$ \sum_{k=0}^m\left(\cos k + i \sin k\right) = \sum_{k=1}^m e^{i k} = \frac{e^{i (m+1)-1}}{e^i-1} $$

but

$$ \mathcal{I}\left(\frac{e^{i (m+1)-1}}{e^i-1}\right) = \frac{1}{\sin(1)}\sin\left(\frac m2\right)\sin\left(\frac{m+1}{2}\right) < 2 $$