Explaining the Limit Comparison Test for Series (1+n *ln(n)) / (n^2 + 5)

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Homework Help Overview

The discussion revolves around the application of the limit comparison test for determining the convergence or divergence of the series (1+n *ln(n)) / (n^2 + 5). Participants are exploring the reasoning behind selecting 1/n as a comparison series.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the validity of using 1/n as a comparison series, particularly in relation to its behavior compared to (ln(n))/n for large n. There is an exploration of the implications of the limit comparison theorem and its conditions.

Discussion Status

Some participants have provided insights into the reasoning behind the choice of 1/n, noting its known divergence and its relationship to the series in question. There is an ongoing exploration of the conditions under which the limit comparison test applies, with no explicit consensus reached.

Contextual Notes

Participants are considering the implications of the limit comparison theorem, particularly regarding the behavior of the series as n approaches infinity and the assumptions made about the divergence of the series being compared.

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In my textbook there is an example that shows determination whether a series converges or diverges using the limit comparison test.
"
The series is (1+n *ln(n)) / (n^2 + 5)
For n large, we expect an to behave like (n*ln(n))/n^2 = (ln(n))/n, which is greater than 1/n for n>= 3, so we take bn = 1/n.
"

My question is why can I use 1/n as an comparison series, I guess it's because its smaller than (ln(n))/n for large n but I'm not sure how that should validate the choice.

Could someone please explain to me?
 
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Swatch said:
In my textbook there is an example that shows determination whether a series converges or diverges using the limit comparison test.
"
The series is (1+n *ln(n)) / (n^2 + 5)
For n large, we expect an to behave like (n*ln(n))/n^2 = (ln(n))/n, which is greater than 1/n for n>= 3, so we take bn = 1/n.
"
My question is why can I use 1/n as an comparison series, I guess it's because its smaller than (ln(n))/n for large n but I'm not sure how that should validate the choice.
Could someone please explain to me?

"I guess it's because its smaller than (ln(n))/n for large n "

Yes, that's the whole point- together with the fact that \Sigma \frac{1}{n} does not converge. The "Limit Comparison Theorem" says that if a_n\leq b_n for large enough n, and \Sigma a_n does not converge, then \Sigma b_n does not converge.
 
they used 1/n because they had a hunch that the series diverges. so when you do the limit comparison test you calculate lim an/bn and if the answer is a finite number greater than zero then either both of your series converge or both diverge. since 1/n is a known divergent series the series you compared it to must also diverge.
it is good test to know for an exam because it is so easy and doesn't require a bunch of inequalities and lots of thinking. if you suspesct your series an to converge then just pick a series bn that converges and calculate lim an/bn if its finite and bigger than zero then you are done.
 
Thanks.:smile:
 

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