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