For an application I'm working on, I need to know the value of the following expression:
$\sum_{i=0}^\infty$ $a^i (2 b-1)\frac{(\frac{b}{(1-b)})^i}{(b-1)} $
(sorry it's not texed). According to WolframAlpha, it converges for |a| < |1/b - 1|, but WA does not give me what it converges to. Does anyone know? (It can probably be found easily with Mathematica)
For any $N\geq 0$, $$ \sum_{i=0}^N \frac{a^i (2 b-1) \left(\frac{b}{1-b}\right)^i}{b-1} =\frac{2 b-1}{b-1}\sum_{i=0}^N a^i \left(\frac{b}{1-b}\right)^i =\frac{2 b-1}{b-1}\sum_{i=0}^N r^i $$ for $r\stackrel{\rm def}{=} a \left(\frac{b}{1-b}\right)$. It will (be defined and) converge when both (i) $b\neq 1$ and (ii) $\left \lvert a \left(\frac{b}{1-b}\right) \right\rvert < 1$, and then the sum will be $$ \frac{2 b-1}{b-1}\cdot \frac{1}{1-r} = \frac{2 b-1}{b-1} \frac{1}{1-a \left(\frac{b}{1-b}\right)} = \frac{2 b-1}{b-1+a b}. $$ from the formula of a geometric series.
