Finance-Theory · Unit 3 · Video 6 · Interactive Practice

The Growing Perpetuity: C/(r − g), and Why Growth Can't Outrun the Discount Rate

IKey Formulas

Formula	Name	What it says
$\text{PV} = \frac{C}{(1+r)} + \frac{C(1+g)}{(1+r)^2} + \frac{C(1+g)^2}{(1+r)^3} + \cdots$	The growing stream, discounted	The first payment $C$ arrives at date 1; each later payment is $(1+g)$ times the one before
$\text{PV} = \frac{C}{r-g}, \quad r > g$	Growing perpetuity	The whole series in one line; $C$ is the date-1 payment, and nothing is paid today
$\frac{1+g}{1+r}$	Ratio of consecutive discounted payments	Below 1 the terms shrink and the sum is finite; at 1 or above they never shrink and it is infinite
$g = 0: \quad \frac{C}{r-0} = \frac{C}{r}$	Level perpetuity	The same formula with no growth — the level perpetuity from earlier in this unit

Key Insight: Growth enters as a subtraction in the denominator, so it raises the value: at $r = 10\%$ a claim paying $100 next year is worth $1,000 level and $2,000 growing at $g = 5\%$ . The condition $r > g$ is not fine print. At $g = r$ the growth exactly offsets the discounting, every discounted payment is the same positive amount, and the sum is infinite; beyond it the formula prints a negative number that no claim to rising payments can be worth.

IIThe Race Between Growth and Discounting

Each payment is multiplied by $1+g$ ; each extra year of discounting divides by $1+r$ . Which wins?

Growth rate g 5% Dates added 4

💡 Only the first sixteen dates are drawn — while $g < r$ the terms after them close the gap to the dashed $C/(r-g)$ line; at $g \ge r$ nothing bounds the running total.

IIISumming the Series

Multiplying by $\frac{1+r}{1+g}$ shifts every term one place left, so all but one cancel.

Step 1

IVValue Against the Growth Rate

As $g$ climbs toward $r$ , the denominator $r-g$ shrinks to nothing, and the value does not stop.

💡 The formula applies one growth rate to every year from here to infinity, so an economy that grows faster than $r$ for fifteen years, as China's did to 2008, is no counterexample.

VQuiz Questions

Problem 1 · Price a Growing Claim

Given: A pipeline contract pays $200 one year from now, and every later payment is 3% larger than the one before, forever. The discount rate is 8%. Find the value of the contract today.

✅ Correct! The denominator is

r - g = 8\% - 3\% = 5\%

, so the contract is worth twenty times its first payment.

❌ That is $C/r$ — the level perpetuity. Growth belongs in the denominator: the divisor is

r - g

, not

r

❌ Check the sign in the denominator. You divided by

r + g = 11\%

. A claim whose payments grow must be worth more than the level claim, not less.

❌ Close — one growth step too many. The

C

C/(r-g)

is already the date-1 payment, so it is not grown again before dividing.

❌ Not quite. Use

\text{PV} = C/(r-g)

with

C = 200

r = 0.08

and

g = 0.03

Show solution

The payments are $200 at date 1, then $200 × 1.03, then $200 × 1.03², and so on forever, so the contract is a growing perpetuity with $C = 200$ , $g = 0.03$ and $r = 0.08$ . Since $r > g$ , the formula applies:

\text{PV} = \frac{C}{r-g} = \frac{200}{0.08 - 0.03} = \frac{200}{0.05} = 4{,}000

The contract is worth $4,000. A level claim paying $200 forever would be worth $\frac{200}{0.08} = 2{,}500$ , so three percent of annual growth, applied forever, is worth $1,500 more than the level stream that starts at the same payment.

Problem 2 · When the Formula Stops Applying

Given: A share is expected to pay a dividend of $3 one year from now, and the dividend is expected to grow 6% a year forever. Investors require a 4% return. Find the value of the share.

✅ Correct! Each discounted dividend is

\frac{1.06}{1.04} \approx 1.019

times the one before, so the terms grow and the running total passes any figure.

❌ Check the subtraction.

r - g = 0.04 - 0.06 = -0.02

, so the formula returns a negative number here — and that number is not the answer either.

❌ That is what the formula prints, but it is not a price. A claim to payments that rise every year cannot be worth less than nothing; the multiply-and-subtract derivation assumed the present value was already finite.

❌ That is $C/g$ . The denominator of the growing perpetuity is

r - g

, and here it is negative.

❌ Not quite. Compare

g

with

r

before reaching for

C/(r-g)

Show solution

Go back to the series the formula is supposed to sum. Each discounted dividend is the one before times

\frac{1+g}{1+r} = \frac{1.06}{1.04} \approx 1.0192 > 1

so the terms are $\frac{3}{1.04} \approx 2.88$ , then $\approx 2.94$ , then $\approx 3.00$ , each larger than the last. Adding infinitely many terms that never shrink gives an infinite sum.

Substituting anyway gives

\frac{C}{r-g} = \frac{3}{0.04 - 0.06} = \frac{3}{-0.02} = -150

which is the algebra breaking, not a valuation: the derivation subtracted one copy of the series from another, and that step is only legitimate when the sum is a finite number. In practice a negative denominator means the inputs are wrong — no dividend grows faster than its required return forever.

Problem 3 · The Terms and Their Sum

Given: A claim pays $50 at date 1, and each later payment is 4% larger than the one before, forever. The discount rate is 9%.

By what factor does each discounted payment change from one date to the next?

What is the whole claim worth today?

✅ Correct! A fixed factor of

0.954 < 1

between consecutive terms is exactly what makes the infinite sum settle at

C/(r-g) = 1{,}000

❌ Check both movements. From one date to the next the payment is multiplied by

1+g

and divided by one more power of

1+r

, so the discounted payment changes by

\frac{1+g}{1+r}

❌ Check the denominator. It is

r - g = 9\% - 4\% = 5\%

— neither

r

alone nor

r + g

, and

C

is not grown before dividing.

Show solution

The date- $t$ payment is $C(1+g)^{t-1}$ and it is discounted $t$ years, so consecutive discounted payments are in the ratio

\frac{C(1+g)^{t}/(1+r)^{t+1}}{C(1+g)^{t-1}/(1+r)^{t}} = \frac{1+g}{1+r} = \frac{1.04}{1.09} \approx 0.9541

Each discounted payment is about 95.4% of the one before — a fixed factor below 1, so the terms shrink geometrically and the sum is finite:

\text{PV} = \frac{C}{r-g} = \frac{50}{0.09 - 0.04} = \frac{50}{0.05} = 1{,}000

Check it against the series. The first discounted payment is $50/1.09 \approx 45.87$ , and a geometric series with first term $a$ and ratio $q$ sums to $a/(1-q)$ :

\frac{45.87}{1 - 0.9541} = \frac{45.87}{0.0459} \approx 1{,}000

So the claim is worth $1,000, twenty times its first payment.

Problem 4 · Two Claims, One Discount Rate

Given: Both claims below are discounted at 10%. Claim A pays $100 a year forever, starting one year from now, with no growth. Claim B pays $60 one year from now, and every later payment is 4% larger than the one before, forever.

What is Claim B worth today?

Which claim is worth more today?

✅ Correct! A smaller first payment and a smaller denominator cancel exactly:

\frac{60}{0.06} = \frac{100}{0.10}

❌ Check the denominator. Growth is subtracted from the discount rate:

r - g = 10\% - 4\% = 6\%

, and

C = 60

is already the date-1 payment.

❌ Value both claims before comparing. A larger first payment and a faster-growing stream pull in opposite directions, so the ranking cannot be read off either feature alone.

Show solution

Claim A is a level perpetuity and Claim B a growing one, both at $r = 0.10$ :

\text{A}: \ \frac{100}{0.10} = 1{,}000 \qquad \text{B}: \ \frac{60}{0.10 - 0.04} = \frac{60}{0.06} = 1{,}000

They are worth the same $1,000 today, even though they never pay the same amount in any single year. Claim B pays $60 against A's $100 next year and stays behind for over a decade — $60(1.04)^{13} \approx 99.9$ in year 14 — passing A only in year 15, and then running away from it forever.

Discounting values the whole stream at once: the early shortfall, the late excess and the timing of each are already inside the two numbers. Ranking claims by their first payment, or by whether they grow, is not a valuation.

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