Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a simple question. While doing geometric series: $$\sum_{i=1}^\infty ar^i = \frac{a}{1-r}.$$ But what if I have something like $\sum_{i=1}^\infty iar^{i-1}$? I think its geometric series, please correct me if I am wrong!

share|cite|improve this question
1… should help (only slightly different than yours). There are several other similar posts... – David Mitra Feb 3 '12 at 19:35
up vote 5 down vote accepted

Note: Your first formula is incorrect as written; the sum should go from $0$ to $\infty$ for that result; alternatively, you the result should be $\frac{ar}{1-r}$ (only works if $|r|\lt 1$, though).

No, your second series is not a geometric series, because the ratio of successive terms is not constant.

However: The second series is obtained from the first one by differentiation. You can use the theory of Taylor Series: $$\begin{align*} \frac{a}{1-r} &= \sum_{i=0}^{\infty}ar^i &\text{if }|r|\lt 1\\ \frac{d}{dr}\frac{a}{1-r} &=\frac{d}{dr}\sum_{i=0}^{\infty}ar^i&\text{if }|r|\lt 1\\ \frac{a}{(1-r)^2} &= \sum_{i=0}^{\infty}\frac{d}{dr}ar^i &\text{if }|r|\lt 1\\ \frac{a}{(1-r)^2} &= \sum_{i=0}^{\infty} iar^{i-1}&\text{if }|r|\lt 1\\ \frac{a}{(1-r)^2} &=\sum_{i=1}^{\infty} iar^{i-1} &\text{if }|r|\lt 1 \end{align*}$$

share|cite|improve this answer
Thanks for the correction! – Sadiksha Gautam Feb 3 '12 at 19:41

If you mean

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

this is just the derivative of

$$\sum_{i=0}^\infty r^i=\frac{1}{1-r}$$

and so

$$\sum_{i=1}^\infty i r^{i-1}=\frac{d}{dr}\frac{1}{1-r}=\frac{1}{(1-r)^2}.$$

All this it is true provided $|r|<1$.

share|cite|improve this answer
Thanks that helped :) – Sadiksha Gautam Feb 3 '12 at 19:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.