Series Convergence: Trouble Determining Convergence/Divergence

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Homework Help Overview

The discussion revolves around determining the convergence or divergence of several series, specifically involving terms like sqrt(n/(n^4-2)), sin(pi/x), and sin(x). The subject area includes series convergence tests and properties of trigonometric functions.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore various convergence tests, including the ratio test and comparisons to known convergent series. Questions arise regarding the summation indices and the behavior of the terms as they approach infinity.

Discussion Status

Participants are actively discussing the convergence of the series, with some offering comparisons to other series and questioning the implications of their findings. There is a mix of interpretations and approaches being explored, particularly regarding the first series and its comparison to 1/n^(3/2).

Contextual Notes

There is a noted lack of clarity regarding the summation indices for the series, and some participants express uncertainty about the behavior of the terms involved, particularly for sin(x) as x approaches infinity.

akoska
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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?
 
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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.
 
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
 
The first one, as Dick said, can be compared to 1/n3/2.

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?
 
So

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

Correct?
 
Wait, sqrt(n/(n^4-2)) > 1/n^3/2, right? So it doesn't matter that 1/n^3/2 converges?
 
How about writing something like 2/n^(3/2)>sqrt(n/(n^4-2))?
 
Last edited:

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