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A formula for approximating ln(2) and sums of other alternating series

  1. Mar 4, 2012 #1
    [tex]1 \ - \ \frac{1}{2} \ + \ \frac{1}{3} \ - \ \frac{1}{4} \ + \ ... \ - \ \frac{1}{n - 1} \ + \ \frac{1}{n} \ - \ \frac{1}{2n + 1} \ < \ ln(n),[/tex] where n is a positive odd integer



    I worked this out (rediscovered it) and proved it by induction.

    For example, when n = 71 (summing of 71 terms and that 72nd fraction),

    it gives five correct decimal digits (and six correct decimal digits

    when rounded).


    [tex]But, \ \ without \ the \ \frac{1}{2n + 1} \ term \ to \ be \ subtracted, \ \ it \ gives \ \ 0.7... \
    (zero \ \ correct \ \ decimal \ \ digits).[/tex]
     
  2. jcsd
  3. Mar 4, 2012 #2

    chiro

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    Science Advisor

    Hey checkitagain.

    Have you seen the wikipedia page? If you want to know more about these expansions you need to understand what is known as a series expansion: the taylor series is a good start. Take a look at these links:

    http://en.wikipedia.org/wiki/Natural_logarithm#Derivative
    http://en.wikipedia.org/wiki/Taylor_series

    In terms of convergence properties, that is a whole other kettle of fish. If you want to know about showing convergence for this kind of thing then there are a variety of theorems you can use.

    This might give you a start:

    http://en.wikipedia.org/wiki/Convergence_(mathematics)#Convergence_and_fixed_point
     
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