Increasing Annuities

An increasing annuity-immediate pays 1, 2, 3, ..., n — a stream that steps up by one unit every period, like escalating rent or a growing coupon. Its present value has a clean closed form, and letting it run forever gives a second closed form for the increasing perpetuity.

By the end you'll be able to compute \((Ia)_n = (\ddot a_n - nv^n)/i\), value an increasing perpetuity as \((Ia)_\infty = 1/i + 1/i^2\), and explain why the sum settles at a finite ceiling instead of growing forever.

Predict: doubling the number of payments in an increasing annuity — does its present value more than double? Drag n and check.

Each bar below is one discounted payment: period k pays k, worth \(k \cdot v^k\) today. Drag n and i to reshape the bars, then watch the meter underneath — it fills to the running total \((Ia)_n\), climbing toward the dashed perpetuity ceiling \((Ia)_\infty\) as n grows.

\((Ia)_{10}\) = (ä₁₀ − 10v¹⁰)/i = (7.802 − 5.584)/0.06 = 36.96

Each bar is one discounted payment k·vᵏ for k = 1...n
discounted payment k·vᵏ
Running total \((Ia)_n\) = 36.96 Ceiling \((Ia)_\infty\) = 294.44

Instead of a level stream, the increasing annuity pays more each period — a small change with a surprisingly clean present-value formula, and a perpetuity limit that settles down instead of exploding.

Formal

Write payment k (which arrives at time k) as a stack of k unit "layers": layer j is a level payment of 1 running from time j through time n — a deferred annuity-immediate. Summing all n layers and simplifying gives \((Ia)_n = (\ddot a_n - nv^n)/i\) — note the numerator uses the annuity-due \(\ddot a_n\), not the annuity-immediate \(a_n\) you met earlier, and the \(-nv^n\) term trims the "overshoot" from having summed a level-n annuity for every layer. Let \(n \to \infty\): \(\ddot a_n \to 1/d\) and \(nv^n \to 0\), so \((Ia)_\infty = (1/d)/i = 1/i + 1/i^2\) — a level perpetuity piece \(1/i\) plus an extra \(1/i^2\) from the payments growing without bound.

Applied

Picture a lease that raises the rent by one fixed unit every year — year 1 pays P, year 2 pays 2P, and so on. Instead of valuing n separate escalating cash flows by hand, you scale the increasing-annuity factor: total PV is \(P \cdot (Ia)_n\). The same idea underlies growing endowments and step-up bond coupons — anywhere a payment schedule climbs by a constant amount each period.

Worked example

Payments of 1,000, 2,000, ..., 10,000 at the end of years 1–10, \(i = 6\%\): \(d = 0.06/1.06 = 0.056604\), so \(\ddot a_{10} = (1 - 1.06^{-10})/d = 7.801692\). \(10v^{10} = 10 \cdot 0.558395 = 5.583948\). \((Ia)_{10} = (7.801692 - 5.583948)/0.06 = 36.96240\). PV \(= 1{,}000 \cdot 36.9624 = \) 36,962.40.

Your turn

Same idea, smaller numbers: payments 1, 2, 3, 4, 5 at the end of years 1–5, \(i = 4\%\). \(d = 0.04/1.04 = 0.038462\), so \(\ddot a_5 = (1 - 1.04^{-5})/d = 4.629895\). \(5v^5 = 5 \cdot 0.821927 = 4.109636\). Now finish it: \((Ia)_5 = (4.629895 - 4.109636)/0.04 = \) ____

Reveal the answer

\((Ia)_5 = 0.520259/0.04 = \) 13.0065. Set n = 5 and i = 4% on the slider above — the meter's running-total readout should match.

More info — why the perpetuity limit doesn't explode

It looks like it should explode: the payments 1, 2, 3, ... grow forever, so how can their present value settle at a finite number? Discounting wins the race. Payment k is worth \(k \cdot v^k\) today, and \(v^k\) shrinks exponentially while k only grows linearly — exponential decay always beats linear growth eventually, so the individual discounted terms shrink toward zero and the running total flattens out at \(1/i + 1/i^2\). The same P/Q decomposition extends to any arithmetic step size — see the decreasing-annuity lesson linked below in Dive deeper.

Check your understanding

Question 1 of 4

What is \((Ia)_6\) — the present value of payments 1, 2, ..., 6 at the end of years 1–6, if \(i = 5\%\)?

Question 2 of 4

An increasing perpetuity pays 1, 2, 3, ... forever at \(i = 10\%\). Its value is \((Ia)_\infty = 1/i + 1/i^2\). What is this value, and what would you get if you forgot the \(1/i\) term?

Question 3 of 4

\((Ia)_n = (\ddot a_n - nv^n)/i\) uses \(\ddot a_n\), the annuity-due value, in its numerator — not the annuity-immediate value \(a_n = (1-v^n)/i\) you met earlier. Since \(\ddot a_n = (1+i)\cdot a_n > a_n\), what happens if you mistakenly substitute \(a_n\) for \(\ddot a_n\)?

Question 4 of 4

Predict, then reason it through: as \(n\) grows very large in an increasing annuity-immediate at a fixed rate \(i > 0\), \((Ia)_n\) approaches the finite ceiling \((Ia)_\infty = 1/i + 1/i^2\). What does that imply about doubling \(n\) from an already-large value?

Recap

  • Increasing annuity-immediate pays 1, 2, ..., n; its present value is \((Ia)_n = (\ddot a_n - nv^n)/i\) — the numerator uses the annuity-due \(\ddot a_n\), not \(a_n\).
  • The increasing perpetuity (payments 1, 2, 3, ... forever) has PV \((Ia)_\infty = 1/i + 1/i^2\): a level-perpetuity piece plus an extra piece from the unbounded growth.
  • Exponential discounting beats linear payment growth, so \((Ia)_n\) converges to a finite ceiling instead of growing without bound as n increases.
  • If the step size is P instead of 1, the stream P, 2P, ... values as \(P \cdot (Ia)_n\).

Dive deeper

Sources

  • Increasing Annuities