How to determine the nature of a series

Question

Consider the following series :$$ \sum\limits_{n = 1}^\infty {\frac{{\sin (n)\cos (\frac{1}{n})}}{{\sqrt {n + 2} }}} $$

We want to determine if the series diverges or not.

I tried so far using the small-angle approximation for the cosine, but it turned even messier. Is this problem as tough as I see it, or is it there something obvious that I'm overlooking ? If anyone gets any ideas towards a solution I would be most welcome!

What is your goal here? Are you trying to sum the series or just establish convergence? — Spencer, Oct 31 '14 at 00:19

score 2 · Answer 1 · edited Apr 13 '17 at 12:21

2

The partial sums of $\sin(n)$ are bounded. The sequence $\cos(1/n)/\sqrt{n+2}$ goes to zero as $n\rightarrow \infty$. A more general version of the alternating series test then tells us that this series converges conditionally.

edited Apr 13 '17 at 12:21

Community

1

answered Oct 31 '14 at 01:42

Spencer

12,271

Boundedness of partial sums of $\sin n$ needs some clarification, but +1. – Przemysław Scherwentke Oct 31 '14 at 01:48
Agreed, I didn't want to get into the details. There is enough here that the OP or someone else can fill in the rest. – Spencer Oct 31 '14 at 01:54

score 0 · Answer 2 · answered Oct 31 '14 at 00:40

Well the small angle approximation gives $\cos{\frac{1}{n}} = 1 + O(\frac{1}{n^2})$ so $$\frac{\sin{n}\cos{n}}{\sqrt{n+2}} = \frac{\sin{n}}{\sqrt{n+2}} + O\left(\frac{1}{n^2}\right)$$ Since the series of $O(\frac{1}{n^2})$ is absolutely convergent, it only remains to study $$ \sum \frac{\sin{n}}{\sqrt{n+2}} $$ Now it is time to use some standard techniques/criteria...

How to determine the nature of a series

2 Answers2