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

  • Thread starter Swatch
  • Start date
  • Tags
    Textbook
In summary, the conversation discusses the use of the limit comparison test to determine whether a series converges or diverges. The example in the textbook involves a series (1+n*ln(n))/(n^2+5) and for large values of n, the series can be approximated by (ln(n))/n, which is greater than 1/n. This leads to the choice of bn=1/n as a comparison series, as it is smaller than (ln(n))/n. The conversation also mentions the Limit Comparison Theorem, which states that if a series an is smaller than a known divergent series bn, then an must also diverge. This test is useful in determining convergence or divergence quickly and efficiently.
  • #1
Swatch
89
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
  • #2
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 [itex]\Sigma \frac{1}{n}[/itex] does not converge. The "Limit Comparison Theorem" says that if [itex]a_n\leq b_n[/itex] for large enough n, and [itex]\Sigma a_n[/itex] does not converge, then [itex]\Sigma b_n[/itex] does not converge.
 
  • #3
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.
 
  • #4
Thanks.:smile:
 

What is the Limit Comparison Test for series?

The Limit Comparison Test is a method used to determine the convergence or divergence of an infinite series. It involves comparing the given series to a known series with known convergence or divergence behavior.

How do you apply the Limit Comparison Test to the series (1+n *ln(n)) / (n^2 + 5)?

To apply the Limit Comparison Test, we first need to choose a known series with known convergence or divergence behavior. In this case, we can choose the series 1/n. Then, we take the limit of the ratio of the given series to the chosen series as n approaches infinity. If the limit is a positive finite number, then both series will have the same convergence or divergence behavior. If the limit is 0 or infinity, then the given series will have a different convergence or divergence behavior than the chosen series.

Why is the series 1/n chosen for the Limit Comparison Test?

The series 1/n is often chosen for the Limit Comparison Test because it is a simple series with known convergence behavior. The series 1/n is known to converge, specifically to the value of 1.

What is the convergence behavior of the series (1+n *ln(n)) / (n^2 + 5)?

Using the Limit Comparison Test with the series 1/n, we can show that the given series (1+n *ln(n)) / (n^2 + 5) also converges, specifically to the value of 0. This means that the series (1+n *ln(n)) / (n^2 + 5) will approach 0 as n approaches infinity.

Can the Limit Comparison Test be used to prove convergence or divergence for all series?

No, the Limit Comparison Test is not a foolproof method for determining convergence or divergence. It can only be used for certain types of series, specifically those that have positive terms and where the limit of the ratio to a chosen series exists and is a positive finite number. In some cases, other tests or methods may need to be used to determine the convergence or divergence of a series.

Similar threads

Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
170
  • Calculus and Beyond Homework Help
Replies
1
Views
232
  • Calculus and Beyond Homework Help
Replies
2
Views
703
  • Calculus and Beyond Homework Help
Replies
17
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
992
Replies
2
Views
845
  • Calculus and Beyond Homework Help
Replies
14
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
14
Views
1K
Back
Top