Convergence of a Series

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  • #1
hpayandah
18
0

Homework Statement



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

Homework Equations




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
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
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. [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
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
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 [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
 

What is convergence of a series?

Convergence of a series refers to the behavior of a series as the number of terms approaches infinity. If the series approaches a finite number as the number of terms increases, it is said to converge. Otherwise, if the series grows without bounds, it is said to diverge.

How do you determine if a series converges or diverges?

To determine the convergence or divergence of a series, you can use various tests such as the comparison test, ratio test, root test, or integral test. These tests compare the given series to a known convergent or divergent series and use properties of the series' terms to determine the behavior of the given series.

What is an example of a convergent series?

An example of a convergent series is the geometric series 1 + 1/2 + 1/4 + 1/8 + ... which approaches a limit of 2 as the number of terms increases. Another example is the telescoping series 1 + 1/2 + 1/3 + 1/4 + ... + 1/n which approaches a limit of ln(2) as the number of terms increases.

What is an example of a divergent series?

An example of a divergent series is the harmonic series 1 + 1/2 + 1/3 + 1/4 + ... + 1/n, which grows without bounds as the number of terms increases. Another example is the alternating harmonic series 1 - 1/2 + 1/3 - 1/4 + ... + (-1)^n/n, which also diverges.

Can a series converge and diverge at the same time?

No, a series cannot converge and diverge at the same time. A series must either approach a finite limit or grow without bounds, but it cannot exhibit both behaviors simultaneously. However, some series may have subseries that converge while others diverge.

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