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 and Convergence Questions

  1. Feb 27, 2010 #1
    1. The problem statement, all variables and given/known data

    What is the limit as n approaches infinity of: (4n+2^n)/(5n+3^n)

    AND

    Does the series from 1 to infinity of (1+n)/sqrt(1+n^6) converge absolutely, conditionally or diverge?

    2. Relevant equations

    L'hopital's rule, convergence tests...

    3. The attempt at a solution

    For the limit, I attempted to use l'hopital's rule but was left stuck with a 2^n*ln2 at the top and a 3^n*ln3 at the bottom.

    For the series question, I had no idea which test to use :S

    Any help would be greatly appreciated!
     
  2. jcsd
  3. Feb 27, 2010 #2

    HallsofIvy

    User Avatar
    Science Advisor

    It shouldn't be hard to see that [itex]2^n[/itex] and [itex]3^n[/itex] will be much larger than 4n and 5n so, for very large n, this is essentially [itex]\frac{2^n}{3^n}= \left(\frac{2}{3}\right)^n[/itex] which, since [itex]\frac{2}{3}< 1[/itex], goes to 0.

    For a more rigorous proof, use L'Hopital's rule.

    Strictly speaking, L'Hopital's rule only applies to limits of functions, but since [itex]f(x_n)[/itex], as [itex]x_n\to a[/itex] must converge to [itex]\lim_{x\to a} f(x)[/itex], we can apply it to sequences as well.

    Here, the derivative of [itex]4n+ 2^n[/itex] is [itex]4+ 2^n ln(2)[/itex] and the derivative of [itex]5n+ 3^n[/itex] is [itex]4+ 3^nln(3)[/itex]. Since
    [tex]\frac{4+ 2^n ln(2)}{5+ 3^n ln(3)} [/tex]
    is still "infinity over infinity", apply L'Hopital a second time, getting
    [tex]\frac{2^n(ln(2))^2}{3^n (n(3))^2}= \frac{ln(2)}{ln(3)}\left(\frac{2}{3}\right)^n[/tex]
    which clearly converges to 0.

    As for [itex]\sum_{n=0}^\infty (1+n)/\sqrt{1+n^6}[/itex], since it involves only positive numbers, there is no question of "conditional convergence"- it either diverges or converges absolutely.

    You should be able to use the comparison test: show that this is less than some (possibly large) number times [itex]n/\sqrt{n^6}= n/n^3= 1/n^2[/itex] and since the series [itex]\sum 1/n^2[/itex] converges (by the integral test), this series converges absolutely.
     
    Last edited by a moderator: Feb 27, 2010
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook