Exponential Sequences and the Number e
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.
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 $\left(1+\frac{1}{m}\right)^m$.
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 $n$:
$c_n = \left( \frac{n+1}{n} \right)^n$
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 $n \to \infty$:
$b_n = \left( 1 + \frac{1}{2n} \right)^{2n}$
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 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 If $|r|<1$, then $r^n\to 0$, so State the limit of the following geometric sequence: $d_n = 5 \cdot \left( \frac{2}{3} \right)^n$ 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? 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 $n \to \infty$).Adding up a geometric sequence
State the limit of the following geometric sequence: $d_n = 5 \cdot \left( \frac{2}{3} \right)^n$
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…
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…
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 $1$).
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 $e^{-1}=\frac{1}{e}$.
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:
$a_n = \left( 1 + \frac{1}{n} \right)^{3n}$
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:
$e_n = \left( 1 - \frac{1}{n} \right)^n$
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 $1^\infty$:
$f_n = \left( 1 + \frac{4}{n} \right)^n$
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 $1$.
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 $n+5$:
$g_n = \left( 1 + \frac{1}{n} \right)^{n+5}$
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:
$a_n = \left( \frac{n+3}{n-1} \right)^{2n}$
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 $1^\infty$:
$b_n = \left( \frac{2n+5}{2n+1} \right)^{3n-2}$
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.
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:
$c_n = \left( \frac{n^2+2n}{n^2-1} \right)^{5n}$
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:
$d_n = \left( \frac{n-2}{n+4} \right)^{\frac{n^2+1}{n}}$
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:
$a_n = \left( \frac{2n^2-3n+1}{2n^2+n-5} \right)^{\frac{n^2+3}{n}}$
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").
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:
$\lim_{n \to \infty} \left( \frac{n+a}{n+1} \right)^{2n} = e^6$
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:
$\lim_{n \to \infty} \left( \frac{an+3}{2n+1} \right)^{bn} = e^{-4}$
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:
$c_n = \left( \sqrt{\frac{n+2}{n-1}} \right)^{3n+1}$
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 $0$.
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.
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 $n \to \infty$)?
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 $n \to \infty$.
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.
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:
$\$5000$ 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 $C'_n = 5000 \cdot (1.04)^n$.
$\$10000$ 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 ($n=4$).
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 $5.13\%$.