A Perpetuity as the Limit of Finite Annuities

From a Finite Annuity to a Perpetuity

A perpetuity can be viewed as the continuation of a level annuity whose payment count keeps increasing. This viewpoint connects a familiar finite present-value formula to the perpetuity formula and, importantly, makes the convergence condition explicit. A long finite annuity may be close in value to a perpetuity, but it is not automatically equal to one.

Assume a constant annual effective interest rate $i$, with discount factor $v=1/(1+i)$. Consider a positive payment $R$ made at the end of each year. The finite annuity pays $R$ at times $1,2,\ldots,n$; its matching perpetuity pays $R$ at every positive integer time. All present values below are measured at time 0.

Definition: The present value of an $n$-payment level annuity-immediate is $R a_{\overline{n}|}$, where $a_{\overline{n}|}=\sum_{k=1}^{n}v^k=(1-v^n)/i$ when $i\ne 0$. For a positive annual effective rate, the matching perpetuity-immediate has present value $R/i$.

The phrase “matching perpetuity” matters: its first payment is at time 1, just as for the finite annuity. If both streams are deferred by the same number of years, both present values receive the same discounting for that deferral. The later examples use this observation without changing the payment timing within either stream.

Why the Limit Equals the Perpetuity Value

The finite annuity contains the first $n$ payments of the perpetuity. Its present value is a finite geometric sum. For $i>0$, the discount factor satisfies $0<v<1$, so the discounted value of the omitted payments shrinks as $n$ increases.

Write the finite sum.
The payment at time $k$ has present value $Rv^k$, so the first $n$ payments have value $R(v+v^2+\cdots+v^n)$.

Use the geometric-sum formula.
For $i\ne0$, the sum is $R(1-v^n)/i$, since $1-v=i/(1+i)$.

Take the limit.
If $i>0$, then $0<v<1$ and $v^n\to0$. Therefore $R(1-v^n)/i\to R/i$.

1-v=1-\frac{1}{1+i}=\frac{i}{1+i}=iv, \qquad \sum_{k=1}^{n}v^k =\frac{v(1-v^n)}{1-v} =\frac{1-v^n}{i}.

\lim_{n\to\infty} R a_{\overline{n}|} = \lim_{n\to\infty} R\frac{1-v^n}{i} = \frac{R}{i} \qquad (i>0).

There is also a direct way to see the size of the gap. The perpetuity consists of the finite annuity plus all payments from time $n+1$ onward. That omitted tail is itself a perpetuity valued one period before its first payment, then discounted back $n$ years.

\frac{R}{i}-R a_{\overline{n}|} = \frac{R}{i}-\frac{R(1-v^n)}{i} = \frac{Rv^n}{i}.

Theorem: For a positive level payment $R$ and a constant annual effective rate $i>0$, the present value of the $n$-payment annuity-immediate increases toward $R/i$ as $n\to\infty$. The difference from the matching perpetuity is exactly $Rv^n/i$, which is positive for every finite $n$.

The positive-interest condition is essential. If $i=0$, then $v=1$ and the $n$-payment value is $nR$, which grows without bound; there is no finite perpetuity value. If $-1<i<0$, then $v>1$, so discounted payment terms grow rather than shrink as they are moved farther into the future. The finite sums do not converge to a finite perpetuity value. Thus, for positive level payments under a constant annual effective rate greater than $-1$, a finite limit exists in this model only when $i>0$.

Worked Examples

Worked Example: Measuring the Gap for a Finite Annuity

A fund pays $1,200 at the end of each year for 8 years. The annual effective interest rate is 5%. Find the present value of the finite annuity, the value of a matching perpetuity, and the difference between them.

Method. Use the finite annuity formula, then use the tail formula to check the difference. The common mistake avoided here is treating the perpetuity value as though it were the value of the 8-payment contract.

Here $R=1{,}200$, $i=0.05$, $n=8$, and $v=1/1.05$. First calculate the discount power:

v^8=\frac{1}{1.05^8}\approx 0.67683936.

The finite annuity factor and its present value are

a_{\overline{8}|} = \frac{1-0.67683936}{0.05} \approx 6.46321280, \qquad 1{,}200(6.46321280)\approx \$7{,}755.86.

The matching perpetuity value is $R/i$:

\frac{1{,}200}{0.05}=\$24{,}000.00.

Check the difference using the omitted-tail formula:

\frac{1{,}200(0.67683936)}{0.05} \approx \$16{,}244.14, \qquad \$7{,}755.86+\$16{,}244.14=\$24{,}000.00.

The check agrees to the cent. The 8-payment annuity is worth less than the perpetuity because it omits all payments after year 8.

Worked Example: Applying the Limit to a Deferred Stream

A contract pays $600 at the end of each year, with its first payment at time 4. Compare the time-0 value of a contract with 6 payments (at times 4 through 9) with the value of a matching perpetuity. The annual effective interest rate is 8%.

Method. The 6 payments at times 4 through 9 are an annuity-immediate deferred 3 years: the ordinary annuity-immediate pattern would begin at time 1, so the value is multiplied by $v^3$. The matching perpetuity is deferred by the same amount. The timing trap is discounting the first payment by $v^4$ in the annuity factor and then discounting it again.

Here $R=600$, $i=0.08$, and $v=1/1.08$. The required powers and annuity factor are

v^3\approx 0.79383224, \qquad v^6\approx 0.63016963, \qquad a_{\overline{6}|} = \frac{1-0.63016963}{0.08} \approx 4.62287963.

Thus the finite contract’s time-0 value is

600v^3a_{\overline{6}|} \approx 600(0.79383224)(4.62287963) \approx \$2{,}201.88.

The matching perpetuity has the same first payment time, so its time-0 value is

600v^3\frac{1}{0.08} \approx 600(0.79383224)(12.5) \approx \$5{,}953.74.

The difference should equal the value of the omitted payments from time 10 onward. There are 6 years of discounting from the annuity’s final payment time 9 back to time 3, and the deferred perpetuity structure gives the equivalent time-0 tail $600v^9/0.08$:

\frac{600v^9}{0.08} \approx \$3{,}751.86, \qquad \$2{,}201.88+\$3{,}751.86=\$5{,}953.74.

The values reconcile to the cent. In particular, the finite value is not simply the perpetuity value: it excludes the payments after time 9.

Worked Example: Finding How Many Payments Are Needed

A payment of $1,000 is made at the end of each year at an annual effective interest rate of 4%. Find the smallest number of payments for which the finite annuity is within $1,000 of its matching perpetuity.

Method. Set the exact tail value no greater than the required tolerance, then solve the resulting inequality for the integer payment count. This avoids relying on a vague rule that a particular number of years is “long enough.”

The perpetuity value is $1{,}000/0.04=\$25{,}000$. For $n$ payments, the gap is $1{,}000v^n/0.04=25{,}000(1.04)^{-n}$. Requiring this gap to be at most $1,000 gives

25{,}000(1.04)^{-n}\le 1{,}000 \quad\Longleftrightarrow\quad (1.04)^n\ge25 \quad\Longleftrightarrow\quad n\ge\frac{\ln 25}{\ln 1.04}\approx82.07.

Since $n$ must be a whole number, the smallest possible count is $n=83$. To check minimality, the gap at 82 payments is approximately $25{,}000(1.04)^{-82}=\$1{,}003.3$, which exceeds $1,000. At 83 payments it is approximately $25{,}000(1.04)^{-83}=\$964.8$, which is within the required amount. The answer is therefore 83 payments.

Common Mistakes and Practical Interpretation

The limit result describes a sequence of values: first value the 1-payment annuity, then the 2-payment annuity, and so on. For positive interest, those values rise toward the perpetuity value. No finite member of that sequence equals the perpetuity value when $R>0$, because the remaining tail $Rv^n/i$ is still positive.

Using the limit formula for a finite contract. The calculation $R/i$ is the matching perpetuity value, not the value of an $n$-payment annuity. For a finite annuity, retain the factor $1-v^n$.
Ignoring the sign of the interest rate. The geometric terms shrink only when $i>0$. The finite-annuity formula alone does not establish convergence to a finite perpetuity.
Changing payment timing unintentionally. The result here uses end-of-year payments. A stream with a different first payment date must be valued with its own timing represented consistently.
Assuming “long” means “close enough.” The exact gap $Rv^n/i$ lets you decide whether the finite value is sufficiently close for a specified purpose.

Key takeaway: Under a constant positive annual effective rate, $R a_{\overline{n}|}$ approaches $R/i$ as the number of level end-of-year payments grows. The remaining gap is $Rv^n/i$; this exact expression both proves convergence and quantifies how close a finite annuity is to its matching perpetuity. This treatment is aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Check Your Understanding

For each question, use the payment timing stated and assume the annual effective interest rate remains constant.

A fund pays $900 at the end of each year for 10 years at 6%. Find the finite annuity’s present value to the nearest cent. (Answer: $6,624.08)
A perpetuity pays $500 at each year-end, and the annual effective rate is 4%. How much less is a 20-payment annuity-immediate worth than the matching perpetuity? (Answer: $5,704.84)
At a 0% annual effective rate, what is the value of 25 end-of-year payments of $300, and does the sequence of values converge to a finite perpetuity value as the payment count grows? (Answer: $7,500; no, the values grow without bound.)
A stream pays $700 at times 6, 7, and onward, at 7% effective annually. Compare the value of its first 9 payments with the matching perpetuity by finding the difference in their time-0 values. (Answer: $700v^{14}/0.07\approx\$3,878.17$, where $v=1/1.07$.)
Payments of $2,000 are made at each year-end at 5%. What is the smallest number of payments for which the finite annuity is within $500 of the matching perpetuity? (Answer: 90 payments.)

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