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

[checkitagain]

1 \ - \ \frac{1}{2} \ + \ \frac{1}{3} \ - \ \frac{1}{4} \ + \ ... \ - \ \frac{1}{n - 1} \ + \ \frac{1}{n} \ - \ \frac{1}{2n + 1} \ < \ ln(n), 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).


But, \ \ without \ the \ \frac{1}{2n + 1} \ term \ to \ be \ subtracted, \ \ it \ gives \ \ 0.7... \ <br /> (zero \ \ correct \ \ decimal \ \ digits).

 

[chiro]

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