Prove that $\sum_{k=1}^\infty k(1-a)^{k-1}a = \frac{1}{a}$ for $0 \lt a \lt 1$

Question

How do you prove that $\sum_{k=1}^\infty k(1-a)^{k-1}a = \frac{1}{a}$ for $0 \lt a \lt 1$

This is the expected value of dice rolls before rolling a 1 with a dice with $\frac{1}{a}$ sides.

Answer 1

No calculus needed:

\begin{align} \sum_{k=1}^\infty k(1-a)^{k-1}a &= \sum_{k=1}^\infty \sum_{j=1}^k (1-a)^{k-1}a \\&= \sum_{j=1}^\infty \sum_{k=j}^\infty (1-a)^{k-1}a \\&= \sum_{j=1}^\infty \frac{(1-a)^{j-1}}{1-(1-a)}a \\&= \sum_{j=1}^\infty (1-a)^{j-1} \\&=\frac1{1-(1-a))} \\&=\frac1{a}. \end{align}

The rearrangements are justified by the absolute convergence of the geometric series in question since $0\lt a \lt 1.$

Answer 2

Hint: divide by $a$ and integrate.

Answer 3

hint: $1+x+x^2+\cdots+ x^k+\cdots = \dfrac{1}{1-x}, |x| < 1$. Differentiate both sides and plug $x = 1-a$ into the equation to get the answer.