omg
Really OMG !!
It seems easy but really i cant test it !!
Am loosing my head :/
Is this one really hard or am forgetting my informations about testing the series for convergence ??!
Isn't it the case that $\displaystyle -1\leq\sin(n)\leq1$ ?
And doesn't this imply that
$\displaystyle \sum_{n=1}^{\infty}\frac{\sin{n}}{n}$ can be directly compared to $\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^n}{n}$ which converges by the alternating series test?
OK $\displaystyle \frac{sin(n)}{n}\geq\frac{-1}{n}$
But what about $\displaystyle \frac{(-1)^n}{n}$ ?
ohhh since $\displaystyle (-1)^n$ = $\displaystyle \pm 1$
Am not sure about inquality, sometimes this bigger and sometimes its smaller.
You are telling me $\displaystyle \frac{sin(n)}{n}\geq\frac{(-1)^n}{n}$ ?
If the smaller converges, You cant decide that convergence of the bigger.
This is what I know.
Or you are telling me $\displaystyle \frac{sin(n)}{n}\leq\frac{(-1)^n}{n}$ ?