I have an infinite geometric series: $$\sum_{i=1}^{\infty} \frac{r^i}{(1+d)^i},$$ where $d$ is a constant. I would like to use the sum of infinite geometric series formula, but I cannot see how to use it because of the $(1+d)^i$ term. How could I calculate this sum?



Note that the summation can be written as

$$\sum_{i = 1}^{\infty}\bigg(\frac{r}{1+d}\bigg)^i = \sum_{i = 1}^{\infty}\bigg(\frac{r}{1+d}\bigg)\cdot\bigg(\frac{r}{1+d}\bigg)^{i-1}$$

and that

$$\sum_{i = 1}^{\infty}u_1\cdot r^{i-1} = \frac{u_1}{1-r}$$

given that $\vert r\vert < 1$.


Express it as $$ \sum_{i=1}^\infty \left( \frac{r}{1+d}\right)^i $$ which is geometric, albeit with a different common ratio.

  • $\begingroup$ Hi. When I tried this, I get based on the formula: $(1+d)/(1+d-r)$. However, my textbook is saying $r/(1+d-r)$?. Could you point out my mistake? Thanks. $\endgroup$ – Thomas Moore Dec 3 '18 at 16:57
  • 2
    $\begingroup$ The first term and the ratio are both equal to $\frac{r}{1+d}$. Use $S_\infty = \frac{u_1}{1-r}$, which simplifies to the correct answer. You took $u_1$ (the first term) to be $1$, which isn’t correct. $\endgroup$ – KM101 Dec 3 '18 at 17:00

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