Series Convergence: Trouble Determining Convergence/Divergence

[akoska]

I'm having trouble determining whether these series converge or diverge.

1. sigma sqrt(n/(n^4-2))

I tried ratio test, but it gave me 1 as the answer (indeterminate)

2. sigma sin (pi/x)

3. sigma sin(x)

I know that sin(x) is bounded...

Any hints?

 

[Dick]

Uh, what are you summing over? In the first one, if you are summing over n its a lot like n^(-3/2). In the others I'm clueless until you illuminate the first point.

 

[akoska]

oh, sorry... first one: sum over n from n=2 to infinity
2. sum over x from x=1 to infinity
3. sum over x from x=0 to infinity

 

[HallsofIvy]

The first one, as Dick said, can be compared to 1/n^3/2.

For the second one, for small \theta, sin(\theta) is approximately \theta so that as x goes to infinity, the terms are approximately \pi/x. Does that series converge?

For the third one, does sin(x) go to 0?

 

[akoska]

So

1. converge
2. no
3. no, sin(x) doesn't go to 0, so the series diverges

Correct?

 

[akoska]

Wait, sqrt(n/(n^4-2)) > 1/n^3/2, right? So it doesn't matter that 1/n^3/2 converges?

 

[Dick]

How about writing something like 2/n^(3/2)>sqrt(n/(n^4-2))?