Convergence of a Series

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  • #1
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Homework Statement



Ʃ[itex]^{∞}_{n=1}[/itex] [itex]\frac{sin(1/n}{\sqrt{n}}[/itex]

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The Attempt at a Solution


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

since lim[itex]_{n→∞}[/itex]=0 therefore the series is convergent

Do you think I did this right?
 

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  • #2
HallsofIvy
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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. [itex]a_n[/itex] going to 0 is a necessary condition for convergence of the series [itex]\sum a_n[/itex], it is not a sufficient condition.

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

For example, [itex]\sum 1/n[/itex] clearly has 1/n going to 0 but the series does not converge.
 
  • #3
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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. [itex]a_n[/itex] going to 0 is a necessary condition for convergence of the series [itex]\sum a_n[/itex], it is not a sufficient condition.

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

For example, [itex]\sum 1/n[/itex] 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?
 
  • #4
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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 [itex]\frac{sin(1/n}{\sqrt{n}}[/itex] < [itex]\frac{1}{\sqrt{n}}[/itex] 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 definiton -2con(1/n) should also converge.
 

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  • #5
Ray Vickson
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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 [itex]\frac{sin(1/n}{\sqrt{n}}[/itex] < [itex]\frac{1}{\sqrt{n}}[/itex] 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 definiton -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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