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

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 $r \textless 1$, 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 $r=\frac{(1+g)}{(1 + d)}$.