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Sum of infinite series

  1. Jan 31, 2010 #1
    I am trying to understand how to calculate the sum of the following
    infinite series, can someone help please:

    (5/7)2 - (5/7)3 + (5/7)4 - (5/7)5 + ...

    The sum of such a series should be given by:

    a / (1-r)

    But the value of a = 0 (the first term = 0), hence my confusion.

    Thanks
     
  2. jcsd
  3. Jan 31, 2010 #2
    The value of a is 1
     
  4. Jan 31, 2010 #3

    Mark44

    Staff: Mentor

    Rewrite the series with (5/7)2 factored out.

    (5/7)2(1 - 5/7 + (5/7)2 -+ ...)
    Now can you see that a = 1 as VeeEight suggested?
     
  5. Feb 1, 2010 #4
    If a = 0, all the terms will be 0.

    Another point, the formula you quoted gives the limit of the sum of powers from 0 to n as n goes to infinity:

    [tex]\frac{1}{1-r} = \sum_{k=0}^{\infty} \left( -\frac{5}{7} \right)^k = 1 -\frac{5}{7} + \left(\frac{5}{7}\right)^2 - \left(\frac{5}{7}\right)^3 + ...[/tex]

    But your sum is missing the first two terms,

    [tex]\left( -\frac{5}{7} \right)^0 = 1, \left( -\frac{5}{7} \right)^1 = -\frac{5}{7}.[/tex]

    So what you have is:

    [tex]\sum_{k=2}^{\infty} \left( -\frac{5}{7} \right)^k = \frac{1}{1-r} - 1 - \left( -\frac{5}{7} \right).[/tex]

    The general formula for a geometric series where the summation index begins at some value, m, not necessarily 0, is

    [tex]\frac{ar^m}{1-r}, |r| < 1.[/tex]

    http://en.wikipedia.org/wiki/Geometric_progression#Geometric_series
     
  6. Feb 1, 2010 #5

    uart

    User Avatar
    Science Advisor

    Astroboy, just use a=25/49 and r=-5/7 in the formula you already posted.
     
  7. Feb 1, 2010 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi astroboy17! :wink:

    A different way of doing it to to say let the sum be S …

    then if you divide S by (5/7)2, you get S + 1 - 5/7 …

    ie (7/5)2S = S + 2/7 …

    carry on from there. :smile:
     
  8. Feb 1, 2010 #7
    Then we can still use the formula

    [tex]\frac{a}{1-r}[/tex]

    if we just redefine a as (-5/7)2, so:

    [tex]\sum_{k=2}^{\infty}\left ( -\frac{5}{7} \right )^k = \sum_{k=0}^{\infty} \left(-\frac{5}{7} \right )^2 \left( -\frac{5}{7} \right )^k[/tex]

    [tex]= \left(-\frac{5}{7} \right )^2 \enspace \frac{\sum_{k=2}^{\infty} \left( -\frac{5}{7} \right )^k}{\left(-\frac{5}{7} \right )^2}[/tex]

    [tex]= \frac{\left( -\frac{5}{7} \right )^2}{1-\left( -\frac{5}{7} \right )}.[/tex]
     
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