I am trying to show that the series $$\sum _{n=1}^{\infty} \sin n$$ diverges but has bounded partial sums.
Plugging in some terms we see that,
$$\sum _{n=1}^{\infty} \sin n = \sin1 +\sin2 +\sin3+...+\sin n$$
My idea is to try and use $$e^{i\theta} = \cos\theta+i\sin\theta$$ $$e^{in}=\cos n+\sin n$$ $$\sin n = Im(e^{in})$$
But how can I use this to show that a finite geometric series won't converge to anything, therefore diverge, but is bounded?
$\sin n$ doesn't tend to $0$ while $n\to\infty$, therefore the sum of the series doesn't converge. To show the first, just note that there exists a constant $\varepsilon>0$ such that one of $|\sin n|, |\sin(n+1)|$ is greater than $\varepsilon$.
But $$\sum_{n=0}^Ne^{in}=\frac{e^{i(N+1)}-1}{e^i-1}$$ The denominator is constant in $N$, and the numerator is bounded by 2, so this series is bounded. Then its imaginary part is also bounded by the same constant.