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Infinite Series

  1. Jan 4, 2009 #1
    1. Problem

    Prove that
    [tex]\sum_{n=0}^{\infty} \frac{a}{k^n} = a\frac{k}{k-1}[/tex]

    2. Relevant equations

    3. The attempt at a solution

    Don't know how to do it at all.
  2. jcsd
  3. Jan 4, 2009 #2
    The infinite sum

    [tex]\sum_{n=0}^\infty ar^n = \frac{a}{1-r}[/tex]

    where |r|<1 (a necessary condition for convergence) is called the geometric series. (Putting r=1/k is your version.) One way to prove convergence and derive the value of the sum is to first derive the finite sum

    [tex]\sum_{n=0}^N ar^n = a\frac{1-r^{N+1}}{1-r}[/tex]

    and then take the limit.

    You can find a simple proof of this latter sum at Wikipedia: http://en.wikipedia.org/wiki/Geometric_progression#Geometric_series

    and (although redundant after having derived the finite sum) proof of the infinite sum there as well: http://en.wikipedia.org/wiki/Geometric_series#Sum .

    I suggest scrolling slowly if you wish to exercise your mind and have a crack at it yourself.
  4. Jan 4, 2009 #3

    Thanks, I will. 7:30am though should sleep soon.
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