1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Limit comparison test

  1. Sep 20, 2010 #1
    1. The problem statement, all variables and given/known data

    show whether [tex]\sum[/tex] (n+5)/(n3-2n+3) is convergant of divergant

    2. Relevant equations

    limit comparison test, lim an/bn = c where c > 0

    3. The attempt at a solution

    an is given
    let bn = n/n3

    so then:

    lim (n+5)/(n3-2n+3)
    n/n3
    = lim (n+5)n3 / (n3 - 2n + 3)n
    = lim (n4 + 5n3) / (n4 - 2n2 +3n)
    = lim (5n3) / (-2n2+3n)

    now as n -> infinity, lim (5n3) / (-2n2+3n) -> -5n/2

    and as n -> inifinity, -5n/2 -> -inifinity... therefore divergant? since c < 0??


    feel free to tell me i have no idea. these things confuse me so much.
     
  2. jcsd
  3. Sep 20, 2010 #2

    Mark44

    Staff: Mentor

    Instead of bn = n/n3, simplify this to 1/n2.

    The limit you should be working with is
    [tex]\lim_{n \to \infty} \frac{\frac{n + 5}{n^3 - 2n + 3}}{\frac{1}{n^2}}[/tex]
     
  4. Sep 20, 2010 #3
    that limit simplifies to

    lim 5n2/(-2n+3)

    so then my previous explantion still stands...

    now as n -> infinity, lim (5n2) / (-2n+3) -> -5n/2

    and as n -> inifinity, -5n/2 -> -inifinity... therefore divergant? since c < 0??

    so i dunno...

    i also tried taking the simplified limit and dividing through by n2 which gives

    lim 5/(-2/n + 3/n2) and i know the limit of 1/n or 1/n2
    and all varieties is 0... so then

    lim 5/(-2/n + 3/n2) = 5/(0-0) = 0... since that is not >0... then it hasnt worked...
     
  5. Sep 20, 2010 #4
    No, that is incorrect. Remember for a rational function f(n), if you let n go to infinity only the terms with the largest exponent on n remain in the numerator and in the denominator.

    For instance, lim (5n^2 + 1000n)/(3n^2 + 40n + 500) is simply lim (5n^2)/(3n^2) = 5/3. Intuitively as n grows beyond all bounds, 5n^2 completely dominates 1000n in the numerator, and similarly 40n + 500 is insignificant compared to 3n^2 when n is approaching infinity.
     
  6. Sep 20, 2010 #5

    Mark44

    Staff: Mentor

    Check your algebra. For large n, the numerator grows at exactly the same rate as the denominator.
     
  7. Sep 20, 2010 #6
    yeah i used that method of thinking to reduce the limit down:

    so then lim 5n2/(-2n+3) becomes lim 5n/-2

    but then i dunno where to go from there..
     
  8. Sep 20, 2010 #7

    Mark44

    Staff: Mentor

    That's still wrong. The limit is a finite constant.
     
  9. Sep 20, 2010 #8
    mark... yeah i did.... it all comes down to a ratio of -2.5n... so the limit still dpeends on n.

    if n = 1000000000 then it comes to -2500000004... i can see the effect here... i just cant summarise it in a way that gives me an appropriate answer...
     
  10. Sep 20, 2010 #9
    the only finite constant here that would make any relevant sense is -5/2...
     
  11. Sep 20, 2010 #10

    Mark44

    Staff: Mentor

    No, it does not depend on n.

    No.

    The numerator is (n + 5)/(n^3 - 2n + 3). The denominator is 1/n^2. For large n, the numerator is roughly n/n^3 or 1/n^2.

    CHECK YOUR ALGEBRA!!!
     
  12. Sep 20, 2010 #11
    (n+5)/(n3-2n+3) is the numerator. for large values of n... this becomes n/n3

    the denominator is 1/n2

    the numerator can simplify down to 1/n2 also.

    so then:

    lim (n+5)/(n3-2n+3) / 1/n2

    becomes lim 1/n2 / 1/n2

    = 1
    since this is greater than 0... both functions converge or diverge... and since 1/n2 converges... so does (n+5)/(n3-2n+3) ??
     
  13. Sep 20, 2010 #12

    Mark44

    Staff: Mentor

    Which is 1/n^2.
    YES!!!
    Yes.
     
  14. Sep 20, 2010 #13
    thank you. sorry for being frustrating.
     
  15. Sep 20, 2010 #14

    Mark44

    Staff: Mentor

    Here are the steps in evaluating that limit.
    [tex]\lim_{n \to \infty} \frac{\frac{n + 5}{n^3 - 2n + 3}}{\frac{1}{n^2}}[/tex]
    [tex]=\lim_{n \to \infty} \frac{n + 5}{n^3 - 2n + 3} \frac{n^2}{1}[/tex]
    [tex]=\lim_{n \to \infty} \frac{n(1 + 5/n)}{n^3(1 - 2/n^2 + 3/n^3} \frac{n^2}{1}[/tex]
    [tex]=\lim_{n \to \infty} \frac{(1 + 5/n)}{(1 - 2/n^2 + 3/n^3} \frac{1}{1}[/tex]
    = 1

    I did quite a bit of cancelling after factoring out n, n^3, and so forth. All of the terms with n to some power in the denominator go to zero when n gets large.
     
  16. Sep 21, 2010 #15
    i see... factorise out the highest exponent that appears in the equation.

    thanks a lot:)
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook