Is the Series ∑ (sin(1/n)/√n) Convergent?

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Homework Help Overview

The discussion revolves around the convergence of the series ∑ (sin(1/n)/√n). Participants explore the behavior of the terms as n approaches infinity and the implications for convergence.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants examine the limit of the terms as n approaches infinity, with one suggesting the use of L'Hôpital's rule. Others question this approach and emphasize that while the limit of the terms going to zero is necessary for convergence, it is not sufficient. The idea of using a limit comparison test is also proposed.

Discussion Status

The discussion is active, with participants providing feedback on each other's reasoning. Some guidance has been offered regarding the use of the p-series for comparison, and there is an acknowledgment that the original proof was insufficient. Multiple interpretations of the problem are being explored.

Contextual Notes

Participants note the behavior of sin(1/n) as n becomes large and discuss the implications of bounding sin(1/n) between -1 and 1. There is also mention of a graph that illustrates the behavior of the terms.

hpayandah
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Homework Statement



Ʃ^{∞}_{n=1} \frac{sin(1/n}{\sqrt{n}}

Homework Equations




The Attempt at a Solution


lim_{n→∞} \frac{sin(1/n}{\sqrt{n}}= lim_{n→∞} \frac{-2cos(1/n)}{n^{3/2}} with l'hospital rule = 0

since lim_{n→∞}=0 therefore the series is convergent

Do you think I did this right?
 
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I can't imagine why you would use L'Hopital for that: sin(2/n) lies between -1 and 1 for all n while the numerator gets larger and larger. Of course, the limit is 0.


But that doesn't help you. a_n going to 0 is a necessary condition for convergence of the series \sum a_n, it is not a sufficient condition.

If a_n does NOT go to 0, then \sum a_n does not converge but there exist series in which a_n converges to 0 but the series \sum a_n does not converge.

For example, \sum 1/n clearly has 1/n going to 0 but the series does not converge.
 
HallsofIvy said:
I can't imagine why you would use L'Hopital for that: sin(2/n) lies between -1 and 1 for all n while the numerator gets larger and larger. Of course, the limit is 0.


But that doesn't help you. a_n going to 0 is a necessary condition for convergence of the series \sum a_n, it is not a sufficient condition.

If a_n does NOT go to 0, then \sum a_n does not converge but there exist series in which a_n converges to 0 but the series \sum a_n does not converge.

For example, \sum 1/n clearly has 1/n going to 0 but the series does not converge.

Consider doing a limit comparison test. What well known limit involving sine might help here?
 
Thank you for quick reply. Okay now I understand that the proof is not sufficient. Can I instead use the fact that you said -1 < sin(1/n) < 1 and the graph shows this:

Graph attached.

can I say since \frac{sin(1/n}{\sqrt{n}} < \frac{1}{\sqrt{n}} then use the p-series to show convergence or divergence.
or I can think of continuing what I was doing and prove that n^(3/2) converges then by definition -2con(1/n) should also converge.
 

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hpayandah said:
Thank you for quick reply. Okay now I understand that the proof is not sufficient. Can I instead use the fact that you said -1 < sin(1/n) < 1 and the graph shows this:

Graph attached.

can I say since \frac{sin(1/n}{\sqrt{n}} < \frac{1}{\sqrt{n}} then use the p-series to show convergence or divergence.
or I can think of continuing what I was doing and prove that n^(3/2) converges then by definition -2con(1/n) should also converge.

You are doing way too much work. When n is large, 1/n is small and sin(1/n) is 1/n + O(1/n^2), so the nth term is 1/n^(3/2) + O(1/n^(5/2)). Convergence is then obvious.

RGV
 

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