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MAth problem (picture)

  1. Sep 3, 2005 #1
    Last edited: Sep 3, 2005
  2. jcsd
  3. Sep 3, 2005 #2


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    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!
    1*2=(1+1)*1 ok



    Now i se it's obvious stupid me :(...
  5. Sep 3, 2005 #4
  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


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    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
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