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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!

J. M. ain't a mathematician
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Sadiksha Gautam
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    http://math.stackexchange.com/questions/30732/how-can-i-evaluate-sum-n-1-infty-frac2n3n1/ should help (only slightly different than yours). There are several other similar posts... – David Mitra Feb 03 '12 at 19:35

2 Answers2

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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: $$\begin{align*} \frac{a}{1-r} &= \sum_{i=0}^{\infty}ar^i &\text{if }|r|\lt 1\\ \frac{d}{dr}\frac{a}{1-r} &=\frac{d}{dr}\sum_{i=0}^{\infty}ar^i&\text{if }|r|\lt 1\\ \frac{a}{(1-r)^2} &= \sum_{i=0}^{\infty}\frac{d}{dr}ar^i &\text{if }|r|\lt 1\\ \frac{a}{(1-r)^2} &= \sum_{i=0}^{\infty} iar^{i-1}&\text{if }|r|\lt 1\\ \frac{a}{(1-r)^2} &=\sum_{i=1}^{\infty} iar^{i-1} &\text{if }|r|\lt 1 \end{align*}$$

Arturo Magidin
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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 $|r|<1$.

Jon
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