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Precalculus · Mathematics
Real Numbers · Sequences

Exponential Sequences and the Number e

Before you start, make sure you know: Algebraic fractions (Algebra 1)

Your savings account pays interest, and the bank offers to compound it more and more often: monthly, daily, every second. Can you earn without limit just by compounding faster? The answer is a very specific ceiling, and that ceiling is one of the most important numbers in math: the number $e$.

The sequence that defines $e$

Evaluate $\left(1+\dfrac{1}{n}\right)^n$ for larger and larger $n$:

$n$ $\left(1+\frac{1}{n}\right)^n$
$1$ $2$
$10$ $2.5937\ldots$
$100$ $2.7048\ldots$
$1{,}000$ $2.7169\ldots$
$10{,}000$ $2.7181\ldots$

The terms keep increasing, but by smaller and smaller steps.

Graph of the sequence (1 + 1/n)^n: the points rise from the axis toward the dashed line at e, with smaller and smaller jumps, never crossing it. e ≈ 2.71828… (1 + 1/n)n increases and approaches e, but never passes it

Use the sequence below to see how the terms approach $e$ without ever exceeding it:

Why does this limit exist at all? The base $1+\frac{1}{n}$ gets closer to $1$, which tries to keep the power small. The exponent $n$ grows forever, which tries to make it large. The number $e$ is the exact balance between those two forces.

The same limit works with other forms of the exponent. For example, $\left(1+\dfrac{1}{2n}\right)^{2n}$ also tends to $e$: let $m = 2n$, and then $m\to\infty$ too, so it is just $\left(1+\frac{1}{m}\right)^m$.

Practice★☆☆Not started
Find the limit of the sequence formed by the quotient of polynomials raised to $n$: $c_n = \left( \frac{n+1}{n} \right)^n$

Find the limit of the sequence formed by the quotient of polynomials raised to $n$:

$c_n = \left( \frac{n+1}{n} \right)^n$

Solve
Practice★☆☆Not started
Determine the value to which the sequence $b_n$ tends as $n \to \infty$: $b_n = \left( 1 + \frac{1}{2n} \right)^{2n}$

Determine the value to which the sequence $b_n$ tends as $n \to \infty$:

$b_n = \left( 1 + \frac{1}{2n} \right)^{2n}$

Solve

Check the base first

Before using anything about $e$, find the limit of the base. Only one case leads to $e$.

The simplest case is a geometric sequence $r^n$: it tends to $\infty$ if $r>1$ and to $0$ if $0

Adding up a geometric sequence

Sometimes you add the terms of a geometric sequence, as with repeated medication doses. The sum of the first $n$ terms (a geometric series) is

$$S_n = a_1 + a_1 r + \dots + a_1 r^{n-1} = a_1\,\frac{1-r^n}{1-r}.$$

If $|r|<1$, then $r^n\to 0$, so

$$\lim_{n\to\infty} S_n = \frac{a_1}{1-r}.$$
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State the limit of the following geometric sequence: $d_n = 5 \cdot \left( \frac{2}{3} \right)^n$

State the limit of the following geometric sequence:

$d_n = 5 \cdot \left( \frac{2}{3} \right)^n$

Solve
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A vehicle loses $15\%$ of its value each year due to depreciation. If its initial price at the dealership is $\$24000$, the value after $n…

A vehicle loses $15\%$ of its value each year due to depreciation. If its initial price at the dealership is $\$24000$, the value after $n$ years is given by the sequence $Vn = 24000 \cdot (0.85)^n$. What will its value be after $5$ years? What happens to its value as the number of years tends to infinity?

n Vₙ ($) 0 2 4 6 8 10 24000 12000
Solve
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A patient receives a dose of $50$ mg of a medication every $24$ hours. Knowing that the body eliminates $40\%$ of the medication present in that time interval, the…

A patient receives a dose of $50$ mg of a medication every $24$ hours. Knowing that the body eliminates $40\%$ of the medication present in that time interval, the accumulated amount in the body just after the $n$-th dose follows the geometric sequence $C_n = 50 + 50 \cdot 0.6 + 50 \cdot 0.6^2 + \dots + 50 \cdot 0.6^{n-1}$. Calculate the long-term accumulated amount (the limit of the sequence when $n \to \infty$).

n Cₙ (mg) Limit = ? 1 3 5 7 9 50 100
Solve

The general formula for $1^\infty$

The same idea appears with many different bases and exponents. One formula turns any $1^\infty$ limit into a simpler limit: exponent times (base minus $1$).

Three-step plan for a 1 to the infinity limit: check the base and exponent, compute L as the limit of b times (a minus 1), and answer e to the L. 1. Check: an → 1 and bn → ∞ 2. Compute L = lim bn·(an − 1) 3. Answer: the limit is e to the L

The same formula handles bases below $1$. For $\left(1-\frac{1}{n}\right)^n$, we get $b_n(a_n-1)=n\cdot\left(-\frac{1}{n}\right)=-1$, so the limit is $e^{-1}=\frac{1}{e}$.

Practice★☆☆Not started
Calculate the limit of the following exponential sequence: $a_n = \left( 1 + \frac{1}{n} \right)^{3n}$

Calculate the limit of the following exponential sequence:

$a_n = \left( 1 + \frac{1}{n} \right)^{3n}$

Solve
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Calculate the limit of the sequence with general term: $e_n = \left( 1 - \frac{1}{n} \right)^n$

Calculate the limit of the sequence with general term:

$e_n = \left( 1 - \frac{1}{n} \right)^n$

Solve
Practice★☆☆Not started
Determine the limit of the following sequence, knowing that it presents an indeterminate form of the type $1^\infty$: $f_n = \left( 1 + \frac{4}{n} \right)^n$

Determine the limit of the following sequence, knowing that it presents an indeterminate form of the type $1^\infty$:

$f_n = \left( 1 + \frac{4}{n} \right)^n$

Solve

Rational bases and linear exponents

Most problems have a base that is a ratio of polynomials and an exponent like $kn$ or $kn+c$. The routine is always the same: subtract $1$ from the base, combine into one fraction (review algebraic fractions if needed), multiply by the exponent, and take the limit using degrees.

A constant added to the exponent never changes the answer when the exponent is $n+c$. For example, $\left(1+\frac{1}{n}\right)^{n+5}$ gives $L=\lim\frac{n+5}{n}=1$, so the limit is still $e$. The extra $5$ factors of the base just multiply the result by something tending to $1$.

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Find the limit of the exponential sequence whose exponent is $n+5$: $g_n = \left( 1 + \frac{1}{n} \right)^{n+5}$

Find the limit of the exponential sequence whose exponent is $n+5$:

$g_n = \left( 1 + \frac{1}{n} \right)^{n+5}$

Solve
Practice★★☆Not started
Calculate the limit of the following sequence, resolving the corresponding indeterminate form: $a_n = \left( \frac{n+3}{n-1} \right)^{2n}$

Calculate the limit of the following sequence, resolving the corresponding indeterminate form:

$a_n = \left( \frac{n+3}{n-1} \right)^{2n}$

Solve
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Determine the limit of the following sequence by resolving the indeterminate form of the type $1^\infty$: $b_n = \left( \frac{2n+5}{2n+1} \right)^{3n-2}$

Determine the limit of the following sequence by resolving the indeterminate form of the type $1^\infty$:

$b_n = \left( \frac{2n+5}{2n+1} \right)^{3n-2}$

Solve

Harder bases and exponents

Sometimes the exponent is itself a rational expression, or the base has degree $2$ or more. The procedure does not change: compute $b_n(a_n-1)$ and simplify with the limit techniques you already know (compare degrees).

Notice how slowly the terms approach $e^2\approx 7.39$. The limit is exact, but convergence can be slow.

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Find the limit of the sequence combining quadratic polynomials in the base: $c_n = \left( \frac{n^2+2n}{n^2-1} \right)^{5n}$

Find the limit of the sequence combining quadratic polynomials in the base:

$c_n = \left( \frac{n^2+2n}{n^2-1} \right)^{5n}$

Solve
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Calculate the limit of the following sequence, which has an algebraic exponent in the form of a fraction: $d_n = \left( \frac{n-2}{n+4} \right)^{\frac{n^2+1}{n}}$

Calculate the limit of the following sequence, which has an algebraic exponent in the form of a fraction:

$d_n = \left( \frac{n-2}{n+4} \right)^{\frac{n^2+1}{n}}$

Solve
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Calculate the limit of the exponential sequence formed by second-degree polynomials in both the base and the exponent:…

Calculate the limit of the exponential sequence formed by second-degree polynomials in both the base and the exponent:

$a_n = \left( \frac{2n^2-3n+1}{2n^2+n-5} \right)^{\frac{n^2+3}{n}}$

Solve

Square roots and unknown parameters

Roots in the base. A square root is a power with exponent $\tfrac12$, so move it into the exponent:

$$\left(\sqrt{x_n}\right)^{b_n}=x_n^{\,b_n/2}.$$

Finding an unknown parameter. Work backward: the limit is $e^L$, so set $L$ equal to the exponent you are given.

If the base is something like $\frac{an+3}{2n+1}$, it tends to $\frac{a}{2}$. A finite nonzero limit such as $e^{-4}$ is only possible when the base tends to exactly $1$, which forces $\frac{a}{2}=1$. Otherwise the limit would be $0$ or $\infty$ (see "Check the base first").

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Find the value of the real parameter $a$ so that the following equality with the number $e$ is satisfied: $\lim_{n \to \infty} \left( \frac{n+a}{n+1} \right)^{2n} = e^6$

Find the value of the real parameter $a$ so that the following equality with the number $e$ is satisfied:

$\lim_{n \to \infty} \left( \frac{n+a}{n+1} \right)^{2n} = e^6$

Solve
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Calculate the value of the parameters $a$ and $b$ so that the following sequence has a limit of $e^{-4}$, knowing that $a$ is a positive real number:…

Calculate the value of the parameters $a$ and $b$ so that the following sequence has a limit of $e^{-4}$, knowing that $a$ is a positive real number:

$\lim_{n \to \infty} \left( \frac{an+3}{2n+1} \right)^{bn} = e^{-4}$

Solve
Practice★★★Not started
Find the limit of the sequence that combines square roots in its base and a linear exponent: $c_n = \left( \sqrt{\frac{n+2}{n-1}} \right)^{3n+1}$

Find the limit of the sequence that combines square roots in its base and a linear exponent:

$c_n = \left( \sqrt{\frac{n+2}{n-1}} \right)^{3n+1}$

Solve

Growth and decay with base $e$

Many natural processes (radioactive decay, cooling, medicine leaving the body) are modeled directly by an exponential with base $e$:

$$a_n = a_0\, e^{kn}.$$

If $k>0$ the quantity grows without bound. If $k<0$ it decays toward $0$.

Two curves: exponential growth with k positive, rising ever more steeply, and exponential decay with k negative, falling toward the horizontal axis. decay (k < 0): aₙ → 0 growth (k > 0): aₙ → ∞

For growth, take bacteria with $P_n=500\,e^{0.2n}$. Here $k=0.2>0$ so $P_n\to+\infty$, and for example $P_{10}=500e^{2}\approx 3{,}695$.

The pollutant problem uses a filtering sequence $C_0\left(1-\frac{2}{n}\right)^n$ whose limit comes from the formula, $C_0e^{-2}$. The "decay by $e^{-k}$" pattern is the same.

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The amount of a radioactive isotope in a sample decreases exponentially according to the sequence $M_n = 200 \cdot e^{-0.1n}$, where $n$ is the number of years elapsed…

The amount of a radioactive isotope in a sample decreases exponentially according to the sequence $M_n = 200 \cdot e^{-0.1n}$, where $n$ is the number of years elapsed and $M_n$ is measured in grams. How many grams will remain in the sample after $20$ years? To what value does the mass tend in the long term (as $n \to \infty$)?

Solve
Practice★★☆Not started
In a thermodynamics experiment, the temperature difference between a metallic body and its surroundings decreases exponentially following the sequence…

In a thermodynamics experiment, the temperature difference between a metallic body and its surroundings decreases exponentially following the sequence $T_n = 80 \cdot e^{-0.05n}$, where $n$ is the minutes elapsed since the start of cooling. Calculate the temperature difference at $30$ minutes and determine the limit of the sequence when $n \to \infty$.

Solve
Practice★★★Not started
The concentration of a pollutant in a lake is naturally reduced through a process that follows the sequence $C_n = C_0 \left( 1 - \frac{2}{n} \right)^n$, where $n$…

The concentration of a pollutant in a lake is naturally reduced through a process that follows the sequence $C_n = C_0 \left( 1 - \frac{2}{n} \right)^n$, where $n$ represents the number of filtering stages applied. If infinite filtering stages are applied, mathematically prove to what exact fraction of the initial concentration $C_0$ the pollutant will be reduced.

Solve

Continuous compounding

If you compound a principal $P$ at annual rate $r$ a total of $n$ times per year, after one year the balance is $P\left(1+\frac{r}{n}\right)^n$. Let $n\to\infty$ (interest added every instant) and $e$ appears.

Try the interactive below and compare how the balance grows as the number of compounding periods increases:

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$\$5000$ are deposited in a financial institution at an annual interest rate of $4\%$. The formula for the final balance after $n…

$\$5000$ are deposited in a financial institution at an annual interest rate of $4\%$. The formula for the final balance after $n$ years with continuous compounding is given by the sequence $Cn = 5000 \cdot e^{0.04n}$. Calculate the accumulated balance after $10$ years and compare it with the balance obtained if the compounding were annual, whose expression is $C'_n = 5000 \cdot (1.04)^n$.

Solve
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$\$10000$ are invested at a nominal annual interest rate of $5\%$. The sequence describing the final balance if the interest is compounded $n$ times a year is $Cn =…

$\$10000$ are invested at a nominal annual interest rate of $5\%$. The sequence describing the final balance if the interest is compounded $n$ times a year is $Cn = 10000 \left( 1 + \frac{0.05}{n} \right)^n$. Calculate the limit of this sequence when $n \to \infty$ to find the final balance under continuous compounding. Compare this result with the balance obtained if compounded quarterly ($n=4$).

Solve

Back to the start: compounding faster does not make your balance explode. With rate $r$, the balance after one year is capped at $P\,e^{r}$. At $5\%$ the effective yield goes from $5\%$ to a ceiling of about $5.13\%$.

Summary

See all the exercises in this section