Year: 1958
Topic: Analysis Algebra

Problem:
What is the smallest amount that may be invested at interest rate $i$, compounded annually, in order that one may withdraw 1 dollar at the end of the first year, 4 dollars at the end of the second year, $\dots$, $n^2$ dollars at the end of the $n$th year, in perpetuity?

Solution:
The present value of one dollar to be paid after $n$ years is $(1 + i)^{-n}$ dollars. Hence the value in dollars of the given annuity is \[ \sum_{n=1}^\infty n^2 (1 + i)^{-n}. \] Since \[ \frac{1}{1 - x} = \sum_{n=0}^\infty x^n, \] we have \[ \frac{x}{(1-x)^2} = x \frac{d}{dx} \left( \frac{1}{1-x} \right) = \sum_{n=1}^\infty n x^n \] and \[ \frac{x + x^2}{(1-x)^3} = x \frac{d}{dx} \left( \frac{x}{(1-x)^2} \right) = \sum_{n=1}^\infty n^2 x^n \] for $|x| < 1$. Putting $x = 1/(1+i)$, we obtain \[ \sum_{n=1}^\infty n^2 (1 + i)^{-n} = \boxed{\frac{(1+i)(2+i)}{i^3}}. \] (At 6 percent interest, the cost of the annuity would be $10,109.26$.)