Why is $\sum_{i=0}^{+\infty} a^{i}i=\frac{a}{(1-a)^{2}}$? I saw this series in some mathematical proofs but I couldn't find why $\sum_{i=0}^{+\infty} a^{i}i=\frac{a}{(1-a)^{2}}$
 A: The series is as follows :-
$$S=a+2a^2+3a^3+.....+=a[1+2a+3a^2+........]\tag{1.}$$
You can see that the series in the brackets is differentiation of the following series:-
$$S_1=a+a^2+a^3+.........\implies S_1=\frac{a}{1-a}$$
i.e. $$S_1'=\frac{S}{a}\implies \frac{S}{a}=\frac{1}{1-a^2}\implies S=\frac{a}{1-a^2}$$
There we go......
A: You can start with the standard finite evaluation:
$$
1+a+a^2+...+a^n=\frac{1-a^{n+1}}{1-a}, \quad a \neq1. \tag1
$$ Then by differentiating $(1)$ we have
$$
1+2a+3a^2+...+na^{n-1}=\frac{1-a^{n+1}}{(1-a)^2}+\frac{-(n+1)a^{n}}{1-a}, \quad |a|<1, \tag2
$$ multiplying by $a$ and by making $n \to +\infty$ in $(2)$, using $|a|<1$, gives 
$$
\sum_{n=0}^{+\infty} na^n=\frac{a}{(1-a)^2}. \tag3
$$
A: Let $s(n)=\displaystyle\sum_{i=1}^na^ii$, then $aS(n)=\displaystyle\sum_{i=1}^na^{i+1}i$ and
\begin{align}
(1-a)S
&=a+a^2+\cdots+a^n+na^{n+1}\\
&=\frac{a-a^{n+1}}{1-a}+na^{n+1}\\
&=\frac{a-a^{n+1}+(1-a)na^{n+1}}{1-a}.
\end{align}
So we conclude that 
$$s(n)=\frac{a-a^{n+1}+(1-a)na^{n+1}}{(1-a)^2}\quad\mbox{for }n\in\mathbb{N},$$
and then $\displaystyle\lim_{n\to\infty}s(n)=\frac{a}{(1-a)^2}$ because
$-1<a<1$.
