Equal payments made at equal intervals, each earning interest for a different length of time, accumulate to a geometric series with r = 1 + i. Discounting the same payments back to today instead gives r = 1/(1 + i), and letting them run for ever is the one case that needs |r| < 1. All three are the same sum formula.
Core ideas
No new mathematics appears in this lesson. What is new is reading a situation and recognising which geometric series it is.
A payment made at the end of period k sits in the account for the remaining n - k periods, so it is worth P(1 + i)ⁿ⁻ᵏ at the end. The last payment earns nothing at all. Listing the payments from the last backwards gives P, P(1 + i), P(1 + i)², … , which is geometric with first term P.
The condition |r| < 1 belongs to the infinite sum and to nothing else. Sₙ = a₁(1 - rⁿ)/(1 - r) needs only r ≠ 1, and an accumulation over n periods is a sum of n terms. Refusing the formula because r > 1 is an overcorrection: it withholds an exact answer to a question that has one.
If payments are monthly, i is a monthly rate and n counts months. Quoting 12% a year and using 1% a month is a convention — the nominal rate convention — and not a mathematical fact: 1% a month compounds to about 12.68% over a year, not 12%. Which convention is in force has to come from the problem.
The material
The accumulation of a stream of payments, the present value of the same stream, and the payments that never stop.
A sequence of equal payments made at equal intervals. In an ordinary annuity each payment falls at the end of its period; in an annuity-due each falls at the beginning. The rate per period i is the interest one period earns, written as a fraction: 10% is i = 1/10.
Everything below is stated for an ordinary annuity, which is the usual convention for a savings plan or a loan repayment.
Take five payments of 1000, one at the end of each year, into an account paying 10% a year. The first payment is deposited at the end of year 1 and then sits there for four more years; the last is deposited at the moment the account is closed and earns nothing.
Reading the list from the last payment backwards is the whole trick. In that order the first term is P and the ratio is 1 + i, and the sum of n such terms is what the sum formula already gives.
n payments of P, one at the end of each period, at rate i per period with i ≠ 0, are worth S = P[(1 + i)ⁿ - 1]/i at the end of the last period.
The payments are worth P, P(1 + i), …, P(1 + i)ⁿ⁻¹ at the close, which is geometric with a₁ = P and r = 1 + i. Since i ≠ 0 we have r ≠ 1, so Sₙ = a₁(1 - rⁿ)/(1 - r) applies.
Substituting, 1 - r = 1 - (1 + i) = -i, so S = P[1 - (1 + i)ⁿ]/(-i). Multiplying the numerator and the denominator by -1 turns that into P[(1 + i)ⁿ - 1]/i, which is the same expression with both signs flipped rather than a different formula. The Sₙ - rSₙ cancellation is doing all the work; only the letters have changed.
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Two checks are free. Putting n = 1 gives S = P[(1 + i) - 1]/i = P, the single payment itself. And at a positive rate every payment is worth at least P at the close, so S cannot come out below nP — here 5000 — which rules out a whole class of arithmetic slips at a glance.
The other question a stream of payments raises is what it is worth now. A payment of P arriving in k periods is worth P/(1 + i)ᵏ today, because that amount, left to compound, becomes P in k periods. Adding those gives a geometric series again, with first term P/(1 + i) and ratio 1/(1 + i).
i = 1/4, n = 2. The payments are worth 100/(5/4) = 80 and 100/(5/4)² = 64 today, totalling 144. The formula gives 100[1 - (4/5)²]/(1/4) = 100(9/25)(4) = 144, the same number. A rate that large is chosen so the arithmetic stays exact; nothing in the derivation cares how big i is.
A perpetuity pays P at the end of every period for ever. Its present value is the infinite series P/(1 + i) + P/(1 + i)² + … , and here the condition finally matters: r = 1/(1 + i), which is less than 1 whenever i > 0. So the sum exists, and it is [P/(1 + i)]/[i/(1 + i)] = P/i.
P/i = 1000/(1/10) = 10000. That number is not a rounding of an infinite total; it is the value the partial sums approach. The first ten payments are worth about 6144.57 today, the first fifty about 9914.81, and the fiftieth payment alone about 8.52 — the later payments contribute less and less, which is why the total stays finite. Those three figures are rounded to the cent; the underlying values are exact fractions.
One practical warning about exactness. (11/10)⁵ is 161051/100000, an exact fraction, and every figure in this lesson is exact. A rate of 7% a year over 30 years is equally exact and far uglier, and it is tempting to round each term to the cent on the way through. Round once, at the end: each of 360 monthly terms rounded to the cent can be half a cent out, and 360 half-cents is 1.80 — small, but enough for two people doing the same calculation to disagree.
Interactive
Each payment grown for its own number of periods, added up and closed-formed
Method
Two decisions come before any formula: what a period is, and which end of the timeline the answer is wanted at.
Monthly payments mean a monthly i and an n counted in months. Convert before substituting, and say which convention you used if the problem quotes an annual rate.
"What will it be worth when I stop paying" is P[(1 + i)ⁿ - 1]/i. "What is it worth now" and "how much can I borrow against it" are P[1 - (1 + i)⁻ⁿ]/i. They differ by a factor of (1 + i)ⁿ.
For an accumulation, a₁ = P and r = 1 + i with the payments listed backwards. For a present value, a₁ = P/(1 + i) and r = 1/(1 + i). Only a perpetuity needs the |r| < 1 test, and it passes it for every positive rate.
At a positive rate an accumulated value is at least nP and a present value is at most nP, the gap being the interest. An answer on the wrong side of nP is a discounting factor applied the wrong way up.
The sum is computed twice: once by valuing each payment and adding the five numbers, once by the closed form. Both are carried through to a figure in cents.
The two computations agree, which checks this one case: P = 1000, i = 1/10, n = 5. The formula holds for every P, every i ≠ 0 and every n because of the cancellation it was derived from, not because of this agreement.
The exponent on the first payment is 4, not 5. It is deposited at the end of the first year, so it earns interest for the four years that remain, and the same n - 1 that appeared in a₁rⁿ⁻¹ is here wearing a calendar. Raising every exponent by one instead gives 6715.61, which is the answer to the annuity-due question — payments at the start of each year — and is exactly 11/10 times this one.
For a faded present-value pass, value three end-of-year payments of 200 today at 25% per year. The supplied timeline decision is to discount, so the first term is 200/(5/4) and the ratio is 4/5. Add the three discounted payments and use the finite geometric formula as a second computation. Check that the result is below the undiscounted total of 600 before opening the quiz.
You should be able to derive P[(1 + i)ⁿ - 1]/i from a₁ = P, r = 1 + i without looking it up, produce a present value from the same series read the other way, and explain why a perpetuity has a finite value while an annuity with r > 1 still has a perfectly good finite sum over n periods.
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