This strange series : $\sum \frac{\sin(1/n)}{\sqrt{n}}$

Question

I'm trying to demonstrate the convergence of the following series: $$\sum \frac{\sin(1/n)}{\sqrt{n}}$$

I know I can prove the convergence by the comparison test. I started like

$$\frac{\sin(1/n)}{\sqrt{n}} \le \frac{1}{\sqrt{n}},$$

but now I'm stuck because $\frac{1}{\sqrt{n}}$ diverges. With which sequence can I compare it to be able to use the comparison test?

Can you give me a hint?

@DanielFischer I tried $n$. But didn't seem to helps me... What can I use. — hlapointe, Oct 29 '15 at 20:52
$n$ is larger than $1$. What sort of bounds for $\sin x$ do you know? — Daniel Fischer, Oct 29 '15 at 20:53
@DanielFischer Now, the only bounds who appear in my head is ${-1; 1}$. — hlapointe, Oct 29 '15 at 20:54
I suppose that any real number greater than $1$ will be a sup. bound of $\sin(1/n)$. — hlapointe, Oct 29 '15 at 20:57

score 1 · Answer 1 · answered Oct 29 '15 at 20:57

1

Hint: $\sin(\frac1n)\leq \frac1n$

answered Oct 29 '15 at 20:57

score 0 · Answer 2 · answered Oct 29 '15 at 20:59

0

Since $\sin(x) < x$ for $0 < x < \pi/2$,

$\sum_{n=1}^m \frac{\sin(1/n)}{\sqrt{n}} < \sum_{n=1}^m \frac{1/n}{\sqrt{n}} \sum_{n=1}^m \frac{1}{n^{3/2}} $

and this converges by lots of tests.

answered Oct 29 '15 at 20:59

marty cohen

2 Answers2