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!

MAth problem (picture)

  1. Sep 3, 2005 #1
    Math problem (picture) (second problem added)

    Can anyone explain this simplification?

    http://img379.imageshack.us/img379/1786/maproblem1bm.jpg [Broken]

    Kindly Paul-MArtin
     
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Sep 3, 2005 #2

    TD

    User Avatar
    Homework Helper

    You probably know that dividing by a fraction is the same as multiplying it with the inverse. So that (a/b)/(c/d) = (a/b)*(d/c). Applying that here gives

    [tex]\frac{{\frac{{3^{k + 1} }}{{\left( {k + 1} \right)!}}}}{{\frac{{3^k }}{{k!}}}} = \frac{{3^{k + 1} }}{{\left( {k + 1} \right)!}}\frac{{k!}}{{3^k }} = \frac{{3 \cdot 3^k \cdot k!}}{{3^k \left( {k + 1} \right) \cdot k!}} = \frac{3}{{\left( {k + 1} \right)}}[/tex]
     
  4. Sep 3, 2005 #3
    Very good, thank you!,

    convincing myself!
    K=1
    1*2=(1+1)*1 ok

    k=2
    1*2*3=(3*1*2)

    k=3
    1*2*3*4=(4*1*2*3)

    Now i se it's obvious stupid me :(...
     
  5. Sep 3, 2005 #4
    I wonder can anyone simplifi this problem? what am i missing?

    http://img55.imageshack.us/img55/1434/maproblem22pi.jpg [Broken]
     
    Last edited by a moderator: May 2, 2017
  6. Sep 3, 2005 #5
    What? That problem doesn't make any sense. Are you sure you've written it down correctly?
     
  7. Sep 3, 2005 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    1. Your picture says [tex]lim_{x->\infty}[/tex] but there is no x in the formula. Did you mean k instead of x?

    2. If the limit you show equals any y, then, by definition, it is not divergent!!

    I think you are referring to the "ratio" test for infinite series:
    If the limit [tex]lim_{k->\infty}\frac{a_{n+1}}{a_n}= y[/tex]
    and y< 1 then the infinite series [tex]\Sigma_{k=1}^{\infty}a_n[/tex]
    converges, if y>1 then it diverges.

    The general proof of that is given in any calculus text. It basically uses y to compare the series to a geometric series.

    If, eventually, [tex]\frac{a_{k+1}}{a_k}< y[/tex] then we can write ak+1< yak< y2ak-1< ....< yka1 so that the series in dominated by the geometric series [tex]\Sigma_{k=1}^{\infty}a_1y^k[/tex] which converges if y< 1.
     
    Last edited: Sep 3, 2005
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: MAth problem (picture)
  1. Math problems? (Replies: 2)

  2. Math Problem (Replies: 2)

  3. Math problem (Replies: 3)

Loading...