Using Limit Comparison Test to Determine Convergence/Divergence

In summary, when using the limit comparison test to determine convergence or divergence of a series, it is important to choose a comparison series (Bn) that is bigger than the original series (An). In the given conversation, the series (Bn) chosen were 1/n^2 and 1/n^3/2, which were found to be suitable for comparison with the original series. It is also important to note that the choice of (Bn) may vary depending on the original series, as seen in the different choices made for (2) and (3) in the conversation.
  • #1
izen
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Homework Statement



Using the limit comparison test to determine convergence or divergence


(1) [itex]\sum^{\infty}_{n=1} \frac{(ln(n))^{3}}{n^{3}}[/itex]


(2) [itex]\sum^{\infty}_{n=1} \frac{1}{\sqrt{n} ln(x)}[/itex]


(3)[itex]\sum^{\infty}_{n=1} \frac{(ln(n))^{2}}{n^{\frac{3}{2}}}[/itex]


Homework Equations





The Attempt at a Solution



I have problem find thing the another series (Bn) compare with An. First I choose Bn when 'n' grows large to represent An

The first question I chose Bn = [itex]\frac{1}{n^{3}}[/itex]
it is the p-series and p>1 so it converges

and then
[itex]lim_{n\rightarrow\infty} \frac{\frac{ln(n)^{3}}{n^{3}}}{\frac{1}{n^{3}}}[/itex]

[itex]lim_{n\rightarrow\infty} ln(n)^{3} = \infty[/itex]

which is wrong so I change to choose Bn = [itex]\frac{1}{n^{2}}[/itex]

[itex]lim_{n\rightarrow\infty} \frac{\frac{ln(n)^{3}}{n^{3}}}{\frac{1}{n^{2}}}[/itex]

[itex]lim_{n\rightarrow\infty} \frac{ln(n)^{3}}{n}= 0[/itex]


So I can conclude that An is convergent by limit comparison

but why I cannot use Bn = [itex]\frac{1}{n^{3}}[/itex] ?

=========================================

(2) and (3) are the same as my first question

(2) [itex]\sum^{\infty}_{n=1} \frac{1}{\sqrt{n} ln(x)}[/itex]

I chose Bn = [itex]\frac{1}{n^{1/2}}[/itex] but after I took the limit of An/Bn. I got 0. In the answer shows that I have to choose Bn = [itex]\frac{1}{n^{}}[/itex]

(3)[itex]\sum^{\infty}_{n=1} \frac{(ln(n))^{2}}{n^{\frac{3}{2}}}[/itex]

I chose Bn = [itex]\frac{1}{n^{3/2}}[/itex] after I took the limit of An/Bn. I got [itex]\infty[/itex]. In the answer shows that I have to choose Bn = [itex]\frac{1}{n^{5/4}}[/itex]

Please give some comments and advices

Thank you
 
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  • #2
izen said:
but why I cannot use Bn = [itex]\frac{1}{n^{3}}[/itex] ?
The idea is to compare your series with a bigger one with known convergence. 1/n^3 is not bigger, therefore it does not work.

The same happens in the other direction, if you want to show divergence.
 
  • #3
Thank you for the comments

mfb said:
The idea is to compare your series with a bigger one with known convergence. 1/n^3 is not bigger, therefore it does not work.

The same happens in the other direction, if you want to show divergence.
[itex]\sum^{\infty}_{n=1} \frac{(ln(n))^{3}}{n^{3}}[/itex]

The series (Bn) that we choose must represent the original series (An) and we recognize converges or diverges from Bn. (We don't know from the original). the problem is...

How do I know that [itex] \frac{1}{n^3} [/itex]is too small [itex]\frac{(ln(n))^{3}}{n^{3}}[/itex] and[itex] \frac{1}{n^2} [/itex]not too big for [itex]\frac{(ln(n))^{3}}{n^{3}}[/itex] and why not choose[itex] \frac{1}{n}[/itex]. still not get it.
 
  • #4
1/n does not give a converging series.
You could choose 1/n3/2 or 1/n5/2 or something similar, if you like, but 1/n2 is easier to study.
 

1. What is the Limit Comparison Test?

The Limit Comparison Test is a mathematical test used to determine the convergence or divergence of an infinite series. It involves comparing the given series to a known series whose convergence or divergence is already known.

2. How do you use the Limit Comparison Test?

To use the Limit Comparison Test, you first need to choose a known series that is similar to the given series. Then, take the limit of the ratio of the two series as n approaches infinity. If the limit is a positive number, both series either converge or diverge. If the limit is 0 or infinity, then the series diverges. If the limit is a finite positive number, then the two series have the same convergence or divergence behavior.

3. What is the difference between using the Limit Comparison Test and the Ratio Test?

The main difference is that the Limit Comparison Test compares the given series to a known series, while the Ratio Test only looks at the ratio of consecutive terms in the given series. The Limit Comparison Test is usually easier to use when the given series is complicated or difficult to analyze using the Ratio Test.

4. Can the Limit Comparison Test be used for both infinite and alternating series?

Yes, the Limit Comparison Test can be used for both infinite and alternating series. However, it is important to note that for alternating series, the limit must be taken of the absolute value of the ratio of the two series.

5. Are there any limitations to using the Limit Comparison Test?

One limitation of the Limit Comparison Test is that it can only be used for series with positive terms. If the given series has negative terms, then the test cannot be applied. Additionally, the choice of the known series used for comparison can greatly affect the results, so it is important to carefully select an appropriate known series.

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