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} &amp; = &amp; 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} &amp; = &amp; d + 1\cdot d^2 + 1\cdot d^3 + 1\cdot d^4 + ... \\
\\
&amp; &amp; \hskip 11pt + 1\cdot d^2 + 1\cdot d^3 + 1\cdot d^4 + ...\\
\\
&amp; &amp; \hskip 45pt + 1\cdot d^3 + 1\cdot d^4 + ... \\
\\
&amp; &amp; ...
\end{eqnarray*}
Every line in this is a geometric sequence, so
\begin{eqnarray*}
\sum_{t=1}^{\infty} t\cdot d^{t} &amp; = &amp; \hskip 7pt \frac{d}{1-d} \\
\\
&amp; &amp; + \frac{d^2}{1-d} \\
\\
&amp; &amp; + \frac{d^3}{1-d} \\
\\
&amp; &amp; ...
\end{eqnarray*}
Which is again a geometric sequence, thus
\begin{eqnarray*}
\sum_{t=1}^{\infty} t\cdot d^{t} &amp; = &amp; \frac{d}{(1-d)^2}.
\end{eqnarray*}
So
$$
\sum_{t=0}^{\infty} a_t \cdot d^t =
\sum_{t=0}^{\infty} a_0 \cdot d^t + \sum_{t=0}^{\infty} b \cdot t \cdot d^t =
\frac{a_0}{1-d} + \frac{b \cdot d}{(1-d)^2}.
$$