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Deriving a geometric series

  1. Feb 26, 2006 #1
    I am trying to derive the geometric series for the following given
    identities,

    [tex]
    \begin{array}{l}
    \frac{1}{{0.99}} = 1.0101010101... \; \; \; {\rm{ (1)}} \\
    [/tex]


    [tex]
    \frac{1}{{0.98}} = 1.0204081632... \; \; \; {\rm{ (2)}} \\
    \end{array}
    [/tex]

    Here is my answer for (1),

    [tex]
    \sum\limits_{n = 1}^\infty {\left( {\frac{1}{{100}}} \right)} ^n + 1
    [/tex]

    Here is my answer for (2),

    [tex]
    \sum\limits_{n = 1}^\infty {\left( {\frac{1}{{50}}} \right)} ^n + 1
    [/tex]

    Are my answers correct? The only way I can get the correct answer is by
    adding 1 onto the series. Is this the correct way represent the series?
     
  2. jcsd
  3. Feb 26, 2006 #2

    0rthodontist

    User Avatar
    Science Advisor

    Yes, the answers are correct. The reason you are adding 1 onto the series is because this formula for the infinite sum of a geometric series
    1/(1-r)
    holds when the first term is 1. 1/.99 can be written as 1/(1-.01) so you have ratio r and first term 1, and the sequence you wrote for the first one reflects that only when you add 1 to it. You could sum from 0 to infinity and remove the 1, instead of from 1 to infinity and adding in the 1 afterwards, if you think it looks neater.
     
  4. Feb 26, 2006 #3
    You can get rid of adding the ones by changing the summation to start from zero instad of one.
     
  5. Feb 26, 2006 #4
    Ok, I see how changing the lower limit to 0 solves the problem of having
    to add 1 to the series.

    Thanks!
     
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