The sum $S=a+a x+a x^{2}+a x^{3}+\cdots$ for $x<1$ can be determined by realizing that $x S=a x+a x^{2}+a x^{3}+\cdots$ and $(1-x) S=a$, so $S=\frac{a}{1-x}$. Using this, we have $\sum_{i=1}^{\infty} \frac{a i}{a^{i}}=$ $a \sum_{i=1}^{\infty} \frac{i}{a^{i}}=a\left[\frac{1}{a}+\frac{2}{a^{2}}+\frac{3}{a^{3}}+\cdots\right]=a\left[\left(\frac{1}{a}+\frac{1}{a^{2}}+\frac{1}{a^{3}}+\cdots\right)+\left(\frac{1}{a^{2}}+\frac{1}{a^{3}}+\frac{1}{a^{4}}+\cdots\right)+\cdots\right]=$ $a\left[\frac{1}{1-a}+\frac{1}{a} \frac{1}{1-a}+\frac{1}{a^{2}} \frac{1}{1-a}+\cdots\right]=\frac{a}{1-a}\left[1+\frac{1}{a}+\frac{1}{a^{2}}+\cdots\right]=\left(\frac{a}{1-a}\right)^{2}$.