Series Comparison Test for Divergence: sin(1/n) vs 1/(1+n)

[Arman777]

Homework Statement

$\sum _{n=0}^{\infty }\:\sin \left(\frac{1}{n}\right)$

Homework Equations

The Attempt at a Solution

Can I try comparison test by
$\left(\frac{1}{1+n}\right)<sin\left(\frac{1}{n}\right)$

since $\left(\frac{1}{1+n}\right)$ diverges also $sin\left(\frac{1}{n}\right)$ diverges ?

 

[Ray Vickson]

Homework Statement

$\sum _{n=0}^{\infty }\:\sin \left(\frac{1}{n}\right)$

Homework Equations

The Attempt at a Solution

Can I try comparison test by
$\left(\frac{1}{1+n}\right)<sin\left(\frac{1}{n}\right)$

since $\left(\frac{1}{1+n}\right)$ diverges also $sin\left(\frac{1}{n}\right)$ diverges ?

What do YOU think?

 

[Arman777]

What do YOU think?

Its true

 

[Ray Vickson]

Its true

Right, and you should be able to explain why that is.

 

[Arman777]

Right, and you should be able to explain why that is.

I already explained it ? Like how

 

[Mark44]

I already explained it ?

No, you didn't explain it -- you just asserted the inequality with no explanation of why it must be true.

$\left(\frac{1}{1+n}\right)<sin\left(\frac{1}{n}\right)$

 

[Arman777]

$\left(\frac{1}{1+n}\right)<sin\left(\frac{1}{n}\right)$ well I am not sure...I put some numbers and tested.

I thought the number $\frac{1}{n}$ and its relationship with $\sin \left(\frac{1}{n}\right)$

Then suddenly I thought $\frac{1}{\:n+1}$

I don't know how to do an exact prove in this case.

 

[Ray Vickson]

$\left(\frac{1}{1+n}\right)<sin\left(\frac{1}{n}\right)$ well I am not sure...I put some numbers and tested.

I thought the number $\frac{1}{n}$ and its relationship with $\sin \left(\frac{1}{n}\right)$

Then suddenly I thought $\frac{1}{\:n+1}$

I don't know how to do an exact prove in this case.

It actually does not matter for small values of $n$. As long as you have $\sin (1/n) > k/n$ for all sufficiently large $n$ and for some positive constant $k$, that is really all you need.

Anyway, that is not really what I was referring to. I was referring to your assertion that $a_n > b_n > 0$ and $\sum b_n$ divergent implies that $\sum a_n$ is also divergent.

 

[vela]

I thought the number $\frac{1}{n}$ and its relationship with $\sin \left(\frac{1}{n}\right)$

Which is that $\sin \frac 1n < \frac 1n$.

Then suddenly I thought $\frac{1}{\:n+1}$

So you have $\frac 1{n+1} < \frac 1n$.

You have two quantities that are less than $1/n$. It's not clear that you can conclude that one is less than the other, i.e., $\frac 1{n+1} < \sin \frac 1n.$

I put some numbers and tested.

This can suggest that perhaps you're on the right track, but as you know, you need to show it for the general case. But it could also be that you're doing things the hard way. Perhaps a different convergence test would work here. You need to experiment with different ideas to learn what works and what doesn't and to develop your intuition in solving these problems.

 

[Arman777]

Yeah I know it was a long shot..there's also limit test which works but It doesn't come to my mind