Does $\sum\limits_{x=1}^\infty\sin(x)$ converge?

Question

I received a task to find out whether the following series converges:

$$\sum_{x=1}^\infty\sin(x)$$

On first look it seems simple, but as I keep thinking about it, there's not a single lemma or criterion that I can use to tackle the problem.

D'alembert ? Doesn't work: The following is meaningless IMHO: $\lim\limits_{x \to \infty}{}\frac{\sin(x+1)}{\sin(x)}$

That series isn't monotonic... you can't understand the rules for when will a member of the series be negative or positive.

All I know is that $\sin(x)$ is blocked between (-1) and 1.
Though it's easy to see that $ \sum\limits_{x=1}^\infty\lvert\sin(x) \rvert$ diverges.

May I use Leibniz formula for $\pi$ in order to construct 2 subseries:

• One that shows that $\sin(x)$ converges to the limit 1
• Another one that shows that $\sin(x)$ converges to the limit 0

And we know that a series can't converge to 2 different numbers, hence it diverges?

Answer 1

The sequence $(\sin n)$ doesn't converge to $0$ so the given series is divergent.

Answer 2

Hint: The partial sums have an explicit form, because there are the imaginary part of some geometric series.

$$ \sum_{k=1}^n e^{ik} = \frac {1-e^{in}}{1-e^i} = \frac {e^{-in/2}-e^{in/2}}{e^{-i/2}-e^{i/2}} \frac{e^{in/2}}{e^{i/2}} = \frac{\sin \frac n2}{\sin \frac 12} e^{i(n-1)/2} $$ so the $n$th partial sum is $$ \frac{1}{\sin \frac 12} \sin \frac n2\sin{\frac{n-1}2} = \frac{1}{2\sin \frac 12} \left(\cos \frac 12 - \cos \frac{2n-1}4\right) $$

From this, you can explicitly see what values are taken by these partial sums.

Answer 3

If $\sin(n) \to 0$, then $|\cos(n)| \to 1$, so $\sum \cos(n)$ diverges, but $\sin(n+1)= \sin(n)\cos(1) + \cos(n)\sin(1)$, so $\sum \sin(1) \cos(n) = \sum \sin(n+1) - \sum \cos(1)\sin(n)$.