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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)
    = 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


    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


    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


    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


    Staff: Mentor

    No, it does not depend on n.


    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.

  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


    Staff: Mentor

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


    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:)
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