Deriving The Perpetuity Formula

Perpetuity refers to an infinite amount of time. In finance, perpetuity is a constant stream of identical cash flows, C, with no end. In this post I explain how one can derive the Perpetuity Formula.

Perpetuity refers to an infinite amount of time (\(\lim_{n\to\infty}\)). In finance, perpetuity is a constant stream of identical cash flows, (\(C\)), with no end.

The present value (\(PV\)) of a security with perpetual cash flows can be determined as:

\(PV = \frac{C}{(1 + d)} + \frac{C}{(1 + d)^2} + \frac{C}{(1 + d)^3} + \dots + \frac{C}{(1 + d)^n} = \frac{C}{d}\)

with \(d\) being the discount rate or cost of capital. Present value just states:

How much money would you need to deposit into an interest earning account (with rate \(d\)) or investment today, in order to get \(C\) amount of money in \(n\) years.

In other words, present value is the result of interest being deducted or discounted from a future amount (compounding in reverse).

So back to our original formula. Why can we rewrite it as follows?

\(PV = \frac{C}{d}\)

If we look at the original formula we can see that it is a geometric series:

\(s = a + ar + ar^2 + ar^3 + \dots\)

with \(a=\frac{C}{(1 + d)}\) and \(r=\frac{1}{(1 + d)}\).

Since for \(n\to\infty\):

\(a + ar + ar^2 + ar^3 + \dots + ar^n = \frac{a}{(1 - r)}\)

we can easily see that:

\(PV = \frac{C}{(1 + d)} + \frac{C}{(1 + d)^2} + \frac{C}{(1 + d)^3} + \dots + \frac{C}{(1 + d)^n}=\frac{\frac{C}{(1 + d)}}{(1 - \frac{1}{(1 + d)})} = \frac{C}{(1+d)(1 - \frac{1}{(1+d)})} = \frac{D}{d}\)

Let us proof that:

\(a + ar + ar^2 + ar^3 + \dots + ar^n = \frac{a}{(1 - r)}\) for \(n\to\infty\)

Let

\(s = a + ar + ar^2 + ar^3 + \dots + ar^{n-1} = \sum\limits_{i=0}^{n-1}ar^i\)

Multiplying \(s\) with \(r\) we get:

\(rs = ar + ar^2 + ar^3 + \dots + r^n\)

Then:

\(s - rs = a - ar^n\)

Solving this for \(s\) we get:

\(s= a\frac{(1 - r^n)}{(1 - r)}\)

Using this we can \(\lim_{n\to\infty} (a + ar + ar^2 + ar^3 + \dots + ar^n)\):

\(\lim_{n\to\infty} (a + ar + ar^2 + ar^3 + \dots + ar^n) = \lim_{n\to\infty}a\frac{1 - r^{n+1}}{1 - r}\)

Above we used \(r^{n+1}\) simply because our formula \(\sum\limits_{i=0}^{n-1}ar^i=a\frac{(1 - r^n)}{(1 - r)}\) is for \(i=0\dots (n-1)\).

For \(r < 1\), which is our case because \(r=\frac{1}{(1 + d)}\) we get:

\(\lim_{n\to\infty}a\frac{(1 - r^{n+1})}{(1 - r)} = \frac{a}{(1-r)}\)

Similarly we can derive the Present Value of Growing Perpetuity where periodic payments grow at a proportionate rate \(g\):

\[PV = \frac{C}{(1 + d)} + \frac{C(1 + g)}{(1 + d)^2} + \frac{C(1 + g)^2}{(1 + d)^3} + \frac{C(1 + g)^3}{(1 + d)^4} + \dots = \frac{C}{(d-g)}\]

which can be rewritten as:

\[PV = \frac{C}{(1 + d)} + \frac{C}{(1 + d)}(\frac{(1 + g)}{(1 + d)})+\frac{C}{(1 + d)}(\frac{(1 + g)}{(1 + d)})^2 + \dots\]

It is simply a geometric series:

\(s = a + ar + ar^2 + ar^3 + \dots\)

with \(a=\frac{C}{(1 + d)}\) and \(r=\frac{(1+g)}{(1 + d)}\).

Written on August 13, 2021