Why doesn't $\sum \frac{sin(\frac{1}{n})}{\sqrt(n)}$ diverge?

Question

Why doesn't $\sum \frac{sin(\frac{1}{n})}{\sqrt(n)}$ diverge?

I know that it converges, I just want to know what i'm doing wrong here. Here's what I did.

$\sum\frac{-1}{\sqrt n} < $ $\sum \frac{sin(\frac{1}{n})}{\sqrt n}$ $\sum<\frac{-1}{\sqrt n }$.

Both $\sum\frac{-1}{\sqrt n}$ and $\sum \frac{-1}{\sqrt n} $ diverge by the p series test since p=0.5<1. Therefore $\sum \frac{sin(\frac{1}{n})}{\sqrt n}$ also diverges. What am I doing wrong here?

Answer 1

For $n$ large, $\sin (1/n) > 0$ and the terms are positive ,and using $\sin (1/n) < 1/n $ we have the given series converges by comparing it with a $p =:3/2$ series.

Answer 2

Your implication is wrong. For example, we know that,$$n^2>\sqrt n\ \forall n\in\mathbb N$$ $$\Longrightarrow\frac{1}{n^2}<\frac{1}{\sqrt n}\ \forall n\in\mathbb N$$ $$\Longrightarrow\frac{-1}{\sqrt n}<\frac{1}{n^2}<\frac{1}{\sqrt n}\ \forall n\in\mathbb N$$ $$\Longrightarrow\sum\frac{-1}{\sqrt n}<\sum\frac{1}{n^2}<\sum\frac{1}{\sqrt n}\ \forall n\in\mathbb N$$ But we know that $\sum\frac{1}{n^2}$ converges, whereas others diverge.

Hope it is helpful