Why is $\sum_{i=0}^{+\infty} a^{i}i=\frac{a}{(1-a)^{2}}$?

Question

I saw this series in some mathematical proofs but I couldn't find why $\sum_{i=0}^{+\infty} a^{i}i=\frac{a}{(1-a)^{2}}$

Answer 1

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......

Answer 2

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

$$ \sum_{n=0}^{+\infty} na^n=\frac{a}{(1-a)^2}. \tag3 $$

Answer 3

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 $-1<a<1$.