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Perpetuities

Payments forever. The present value is finite because discounting beats accumulation, and the formulas are the cleanest in the syllabus.

The formulas

Perpetuity-immediate

$$a_{\overline{\infty}|} = \dfrac{1}{i}$$

Perpetuity-due

$$\ddot{a}_{\overline{\infty}|} = \dfrac{1}{d} = \dfrac{1+i}{i}$$

Increasing perpetuity

$$(Ia)_{\overline{\infty}|} = \dfrac{1}{i} + \dfrac{1}{i^{2}} = \dfrac{1+i}{i^{2}}$$

Geometric (Gordon)

$$PV = \dfrac{1}{i-g}$$

payments growing at rate g < i

Where it comes from

1. Let $n \to \infty$ in $a_{\overline{n}|} = \frac{1-v^n}{i}$. Since $|v| < 1$, $v^n \to 0$ and the value tends to $1/i$.
2. The due version is one period earlier: multiply by $(1+i)$, giving $1/d$.
3. For payments growing by a factor $(1+g)$ each period, the series has ratio $\frac{1+g}{1+i} < 1$ and sums to $\frac{1}{i-g}$.

Worked example

This answer is recomputed from the site’s own interest-theory and probability functions every time the test suite runs, so the page and the mathematics cannot drift apart.

Memory hooks

• 1/i is 'the fund whose interest exactly is the payment'. Everything else is that plus a timing shift.
• Perpetuity-due = perpetuity-immediate + one payment now = 1/i + 1 = 1/d.

Traps

• Using 1/i when the first payment is immediate - that needs 1/d.
• Applying the Gordon form when g ≥ i, where the present value is infinite.

Related

• Annuity-immediate
• Annuity-due
• Geometrically increasing annuities

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