We have that
$$p=\sum_{t=1}^\infty \frac{d}{(1+r)^t}$$
Where d is a constant and r is a percentage. I am trying to deduce the result $d/r$ by the rule of infinite geometric series that states when a is constant and $|x|<1$,
$$\sum_{t=0}^\infty \ ax^t=\frac{a}{1-x}$$
The closest I get by substitutions is to $-d/r$. I would be very grateful if someone could help me. Thank you.
One may write, by a change of index, using the geometric sum evaluation, $$ \begin{align} \sum_{t=1}^\infty \frac{d}{(1+r)^t}&= \frac{d}{(1+r)}\sum_{t=1}^\infty \frac{1}{(1+r)^{t-1}} \\\\&= \frac{d}{(1+r)}\sum_{t=0}^\infty \frac{1}{(1+r)^{t}} \\\\&=\frac{d}{(1+r)} \cdot\frac{1}{1-\frac{1}{(1+r)}} \\\\&=\frac{d}{(1+r)-1} \\\\&=\frac dr. \end{align} $$