Compound Interest and the Number e

Q: What is the compound interest formula?

A = P(1 + r/n)^(nt): principal P, annual rate r as a decimal, n compounding periods per year, and t years.

Q: What is the number e?

The limit of (1 + 1/n)ⁿ as n grows without bound, about 2.71828. It is the growth factor for one unit of time at a 100% rate compounded continuously.

Q: What does compounded continuously mean?

Interest is added at every instant rather than at set times. The balance is A = Pe^(rt).

Q: Does compounding more often make much difference?

Less and less. At 6% for a year, monthly compounding turns $1000 into $1061.68 and continuous compounding into $1061.84.

Q: What is an effective annual rate?

The simple yearly rate that gives the same result as the compounded one. 6% compounded monthly has an effective rate of about 6.17%.

Quick answer

Interest compounded n times a year multiplies a balance by (1 + r/n)^(nt). Compounding more often earns a little more, but the gains level off: as n grows without bound, (1 + 1/n)ⁿ approaches the number e ≈ 2.71828. Compounding continuously gives A = Pe^(rt), the model for anything that grows or decays in proportion to its size at every instant, from bank balances to bacteria to radioactive atoms.

What you'll learn

Use the compound interest formula for any number of periods per year
Explain why more frequent compounding approaches a limit
Define e and use continuous growth, A = Pe^(rt)
Compare offers with effective annual rates

Compound interest

A bank that pays $6\%$ a year, compounded monthly, pays $\tfrac{6\%}{12} = 0.5\%$ at the end of every month, and each month’s interest earns interest in later months. After $t$ years there have been $12t$ payments, each multiplying the balance by $1.005$ . In general, with principal $P$ , annual rate $r$ and $n$ compounding periods a year,

A = P\left(1 + \frac{r}{n}\right)^{nt}

Here is $1000$ dollars at $6\%$ for one year, compounded more and more often:

Compounded	$n$	Balance after 1 year
annually	$1$	$1060.00$ dollars
quarterly	$4$	$1061.36$ dollars
monthly	$12$	$1061.68$ dollars
daily	$365$	$1061.83$ dollars

The balance grows with $n$ , but by less and less each time.

Why the gains level off

Take the simplest case: $100\%$ interest on $1$ dollar for one year, compounded $n$ times. The balance is $\left(1 + \tfrac{1}{n}\right)^n$ .

(1 + 1/n)ⁿ
e ≈ 2.71828

The balance (1 + 1/n)ⁿ as the number of compoundings grows

Annually gives $2$ dollars; twice a year, $2.25$ dollars; monthly, $2.61$ dollars; daily, $2.71$ dollars; every second of the year, $2.71828$ dollars. Each extra compounding pays interest on interest that was itself earned a moment before, a smaller and smaller amount. Compounding more often helps less and less, and the limit of the process is a number, $e$ :

\left(1 + \frac{1}{n}\right)^n \longrightarrow e \approx 2.71828 \quad\text{as } n \text{ grows}

Continuous growth

Compounding “infinitely often”, or continuously, gives the limit of the formula. The same argument with rate $r$ and $t$ years produces

A = Pe^{rt}

Beyond banking, $y = y_0e^{kt}$ describes any quantity whose rate of change is proportional to its size at every moment: a bacteria culture ( $k > 0$ ) or a radioactive sample ( $k < 0$ ). The base $e$ appears because nature does not wait for the end of the month.

Worked examples

Common mistakes

Practice problems

Find the balance on $2000$ dollars at $5\%$ compounded quarterly for $3$ years.

Answer

About $2321.51$ dollars

Full solution

$2000\left(1 + \tfrac{0.05}{4}\right)^{12} = 2000(1.0125)^{12} \approx 2321.51$ .
Find the balance on the same investment compounded continuously.

Answer

About $2323.67$ dollars

Full solution

$2000e^{0.05 \cdot 3} = 2000e^{0.15} \approx 2323.67$ .
Find the effective annual rate of $8\%$ compounded quarterly.

Answer

About $8.24\%$

Full solution

$\left(1 + \tfrac{0.08}{4}\right)^4 = 1.02^4 \approx 1.0824$ .
Find the balance on $800$ dollars at $3\%$ compounded continuously for $7$ years.

Answer

About $986.94$ dollars

Full solution

$800e^{0.03 \cdot 7} = 800e^{0.21} \approx 986.94$ .
Over $10$ years, which grows $1000$ dollars more: $5\%$ compounded annually or $4.9\%$ compounded continuously?

Answer

$4.9\%$ compounded continuously: $1632.32$ dollars against $1628.89$ dollars

Full solution

$1000(1.05)^{10} \approx 1628.89$ and $1000e^{0.49} \approx 1632.32$ .
A $10\,000$ mg sample decays as $10\,000e^{-0.2t}$ , with $t$ in hours. How much remains after $3$ hours?

Answer

About $5488$ mg

Full solution

$10\,000e^{-0.6} \approx 5488.12$ .
How much must be invested now at $4\%$ compounded continuously to have $10\,000$ dollars in $5$ years?

Answer

About $8187.31$ dollars

Full solution

$Pe^{0.2} = 10\,000$ , so $P = 10\,000e^{-0.2} \approx 8187.31$ .
Compare $1000$ dollars at $5\%$ for $20$ years compounded daily and continuously.

Answer

About $2718.10$ dollars and $2718.28$ dollars

Full solution

$1000\left(1 + \tfrac{0.05}{365}\right)^{7300} \approx 2718.10$ and $1000e^{1} \approx 2718.28$ : continuous growth at $5\%$ for $20$ years multiplies by exactly $e$ .
Estimate $e$ with $\left(1 + \tfrac{1}{n}\right)^n$ for $n = 10$ .

Answer

About $2.594$

Full solution

$1.1^{10} \approx 2.5937$ , still well below $e \approx 2.718$ ; the approach is slow.
A student computes $1000$ dollars at $6\%$ compounded monthly for $2$ years as $1000(1.06)^{24}$ . What went wrong?

Hint

What does each monthly payment multiply the balance by?

Answer

The rate per month is $\tfrac{0.06}{12} = 0.005$ . The balance is $1000(1.005)^{24} \approx 1127.16$ dollars.

Full solution

The student’s $1.06^{24}$ applies a full year’s interest every month, giving about $4048.93$ dollars, far too much. Each of the $24$ months multiplies by $1.005$ .

Frequently asked questions

What is the compound interest formula?

A = P(1 + r/n)^(nt): principal P, annual rate r as a decimal, n compounding periods per year, and t years.

What is the number e?