Series with increasing factor inside the summation 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! 
 A: 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:
$$\begin{align*}
\frac{a}{1-r} &= \sum_{i=0}^{\infty}ar^i &\text{if }|r|\lt 1\\
\frac{d}{dr}\frac{a}{1-r} &=\frac{d}{dr}\sum_{i=0}^{\infty}ar^i&\text{if }|r|\lt 1\\
\frac{a}{(1-r)^2} &= \sum_{i=0}^{\infty}\frac{d}{dr}ar^i &\text{if }|r|\lt 1\\
\frac{a}{(1-r)^2} &= \sum_{i=0}^{\infty} iar^{i-1}&\text{if }|r|\lt 1\\
\frac{a}{(1-r)^2} &=\sum_{i=1}^{\infty} iar^{i-1} &\text{if }|r|\lt 1
\end{align*}$$
A: 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 $|r|<1$.
