Exploring the Relationship Between a/b and Geometric Series

In summary, a probability problem led to the discovery of the relationship $$\frac a b = \frac a {(b-1)} - \frac a {{(b-1)}^2} + \frac a {{(b-1)}^3} - \frac a {{(b-1)}^4} + ~...$$ which can be simplified to a geometric series. The Cesaro sum of this series is 1/2, but it only converges for values of b outside the range of 0 to 2. Further research on similar series may lead to interesting results.
  • #1
Jehannum
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While working on a probability problem I accidentally found this relationship:

$$\frac a b = \frac a {(b-1)} - \frac a {{(b-1)}^2} + \frac a {{(b-1)}^3} - \frac a {{(b-1)}^4} + ~...$$
I have done a bit of work on it myself, and have tried to research similar series. It seems to lead to some interesting results. For example, when a = 1 and b = 2 it doesn't work because you get 1 - 1 + 1 - 1 + 1 ... but it's interesting that the Cesaro sum of this series is 1/2.

Can anyone provide links or information on anything relevant?
 
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  • #3
Jehannum said:
While working on a probability problem I accidentally found this relationship:

$$\frac a b = \frac a {(b-1)} - \frac a {{(b-1)}^2} + \frac a {{(b-1)}^3} - \frac a {{(b-1)}^4} + ~...$$
I have done a bit of work on it myself, and have tried to research similar series. It seems to lead to some interesting results. For example, when a = 1 and b = 2 it doesn't work because you get 1 - 1 + 1 - 1 + 1 ... but it's interesting that the Cesaro sum of this series is 1/2.

Can anyone provide links or information on anything relevant?
Taking out the common factor of ##a## and letting ##x = \frac{1}{b - 1}##, you have a geometric series:
$$S = x - x^2 + x^3 - x^4 + \dots$$This converges for ##|x| < 1## to ##S = \frac{x}{1+ x}##, and a bit of algebra shows that indeed:$$\frac{x}{1+ x} = \frac 1 b$$And ##|x| < 1## implies ##b < 0## or ##b > 2##. In particular, this series does not converge for ##b = 2##.
 
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