# How to find the present value of a perpetuity with a non-constant growth rate?

I've got a perpetuity problem where an organization pays out 50 equal valued grants each year in perpetuity, adding an additional 5 grants each year (i.e. 55 in year 2, 60 in year 3, etc.).

We've only been shown how to do perpetuity problems with a constant growth rate (e.g. the payout increases by 5% each year), so I'm not sure how to deal with this one.

Anyone have any insight?

Basically you want to calculate $$\sum_{t=0}^{\infty} a_t \cdot d^t$$ where $d$ is the discount factor and where the value paid yearly increases linearly, that is $$a_t = a_0 + b \cdot t.$$ With some rearrangements $$\sum_{t=0}^{\infty} a_t \cdot d^t = \sum_{t=0}^{\infty} a_0 \cdot d^t + \sum_{t=0}^{\infty} b \cdot t \cdot d^t.$$ The first half of this is a simple geometric sequence like the present value of a perpetuity, so $$\sum_{t=0}^{\infty} a_0 \cdot d^t = \frac{a_0}{1-d}.$$ To calculate the other sum we'll use a trick. $$\sum_{t=0}^{\infty} b \cdot t \cdot d^t = b \cdot \sum_{t=1}^{\infty} t \cdot d^t.$$ Then write \begin{eqnarray*} \sum_{t=1}^{\infty} t\cdot d^{t} & = & d + 2\cdot d^2 + 3\cdot d^3 + 4\cdot d^4 + ... \end{eqnarray*} You can rearrange this (because of absolute convergence) to \begin{eqnarray*} \sum_{t=1}^{\infty} t\cdot d^{t} & = & d + 1\cdot d^2 + 1\cdot d^3 + 1\cdot d^4 + ... \\ \\ & & \hskip 11pt + 1\cdot d^2 + 1\cdot d^3 + 1\cdot d^4 + ...\\ \\ & & \hskip 45pt + 1\cdot d^3 + 1\cdot d^4 + ... \\ \\ & & ... \end{eqnarray*} Every line in this is a geometric sequence, so \begin{eqnarray*} \sum_{t=1}^{\infty} t\cdot d^{t} & = & \hskip 7pt \frac{d}{1-d} \\ \\ & & + \frac{d^2}{1-d} \\ \\ & & + \frac{d^3}{1-d} \\ \\ & & ... \end{eqnarray*} Which is again a geometric sequence, thus \begin{eqnarray*} \sum_{t=1}^{\infty} t\cdot d^{t} & = & \frac{d}{(1-d)^2}. \end{eqnarray*} So $$\sum_{t=0}^{\infty} a_t \cdot d^t = \sum_{t=0}^{\infty} a_0 \cdot d^t + \sum_{t=0}^{\infty} b \cdot t \cdot d^t = \frac{a_0}{1-d} + \frac{b \cdot d}{(1-d)^2}.$$