Undergrad How Does the A/B Relationship Connect to Geometric Series?

[Jehannum]

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?

 

[fresh_42]

https://www.wolframalpha.com/input?i=sum++(k=1+to+infinity)+(-1)^k+(1/(b-1)^k)

 

[PeroK]

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 $b = 2$.