Determine convergence or divergence $\sum_1^\infty \sin \frac{1}{n}$ $$\sum_1^\infty \sin \frac{1}{n}$$
So now I konw that to evaluate this I can just look at the limit as it reaches infinity. I see that it would result in 1 over 0, but it approaches 0 so I could that that it approaches sin0 which is 0 so doesn't this converge?
 A: Hint: Show that for $0<x\le 1$ we have $x<2\sin x.$ Consequently, for integers $n\ge 1,$ we have $$\frac1n<2\sin\frac1n,$$ so $$\frac12\cdot\frac1n<\frac12\cdot2\sin\frac1n=\sin\frac1n.$$ Use Comparison test with (a multiple of) the harmonic series.
A: [I think you are referring to the divergence test in your post. I hope to clarify the source of confusion, but this is too long for a comment.]
Just to clarify the divergence test:
If $\lim_{n\to \infty} a_n \neq 0$, then $\sum_{n = 1}^\infty$ diverges.
Put differently (it it's contrapositive), if $\sum_{n = 0}^\infty a_n\;$ converges, then $\lim_{n\to \infty} a_n= 0$.
The converse is not necessarily true. For example, $a_n = \dfrac 1n$. It is certainly true that $\lim_{n\to \infty} \dfrac 1n = 0$. However, $$\sum_{n = 1}^\infty \frac 1n \;\;\text{diverges}$$

Indeed, in this example, the harmonic series $\sum_{n=1}^\infty \dfrac{1}{2n}$ will make for a good comparison, using the comparison test: knowing that $$\sum_{n = 1}^\infty \frac 1{2n}$$ diverges, and knowing that $$\dfrac 1{2n} \leq \sin\frac 1n$$, you can argue that your series must also diverge.
A: Hint: $$x\in (0,1/2)\Rightarrow\sin x > x/2$$
