Why use 1/n as the comparison series for the limit comparison test

  • Thread starter Thread starter Swatch
  • Start date Start date
  • Tags Tags
    Textbook
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
3 replies · 3K views
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 [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.
 
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.