How do I find the infinite series of a trig function that's weird?

Question

The title.

For example:

$$\sum_{n=1}^\infty\sin\left(\frac{1}{n}\right)$$

How would I go about solving this. Does it converge? Does it diverge? What does it converge to?

Answer 1

First, show that

$$\frac{x}{2} \leq \sin(x),\quad {0 \leq x \leq 1}.$$

Now, use this inequality to compare your series with one-half the harmonic series.

Answer 2

The sum diverges. This is not hard to show as for small values of $x$, $$\sin(x) \approx x$$ and that $$\sum_{n=1}^\infty\frac{1}{n}= \infty$$

Answer 3

We have $$\lim_n{\sin{1\over n}\over {1\over n}}=1$$ By the limit comparison test the series $\sum \sin{1\over n}$ is divergent as the series $\sum {1\over n}$ is divergent.