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

  • Thread starter Thread starter Swatch
  • Start date Start date
  • Tags Tags
    Textbook
Click For Summary
SUMMARY

The discussion focuses on the application of the Limit Comparison Test to determine the convergence or divergence of the series (1+n *ln(n)) / (n^2 + 5). For large values of n, the series behaves like (ln(n))/n, which is greater than 1/n for n ≥ 3. The choice of 1/n as a comparison series is validated because it is a known divergent series, and the Limit Comparison Theorem states that if a_n ≤ b_n for sufficiently large n and Σ a_n diverges, then Σ b_n also diverges. This method simplifies the process of evaluating series convergence.

PREREQUISITES
  • Understanding of the Limit Comparison Test in calculus
  • Familiarity with series convergence and divergence concepts
  • Knowledge of logarithmic functions and their behavior as n approaches infinity
  • Basic proficiency in evaluating limits
NEXT STEPS
  • Study the Limit Comparison Test in detail, including its formal proof and applications
  • Explore examples of series that converge and diverge using the Limit Comparison Test
  • Learn about other convergence tests such as the Ratio Test and Root Test
  • Practice evaluating limits of the form lim an/bn for various series
USEFUL FOR

Students and educators in calculus, particularly those focusing on series convergence, as well as anyone preparing for exams that cover series analysis and convergence tests.

Swatch
Messages
88
Reaction score
0
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?
 
Physics news on Phys.org
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:
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
Replies
2
Views
1K
  • · Replies 14 ·
Replies
14
Views
2K