= Solution
The absolutely convergent <geometric series> has sum $\sum_{k=0}^{\infty}r^k=(1-r)^{-1}$ for $0<r<1$. Its <Cauchy product> with itself is
$$
\left(\sum_{k=0}^{\infty}r^k\right)^2
=\sum_{m=0}^{\infty}\left(\sum_{k=0}^{m}1\right)r^m
=\sum_{m=0}^{\infty}(m+1)r^m.
$$
There are exactly $m+1$ pairs of nonnegative integers whose sum is $m$. Relabeling $j=m+1$ proves
$$
\boxed{\sum_{j=1}^{\infty}j r^{j-1}=\frac1{(1-r)^2}.}
$$
Absolute convergence justifies the multiplication and regrouping; no differentiation of an infinite series is needed.
Back to article page
