Exponential sequences. The number e
If you invest money and the bank pays you interest more and more times a year (monthly, daily, every second…), how much more can you earn? The answer has a very specific limit, and that limit is one of the most important numbers in mathematics: the number $e$.
The sequence that defines the number $e$
Calculate $\left(1+\dfrac{1}{n}\right)^n$ for larger and larger $n$: $n=1\to2$; $n=10\to2{,}5937\ldots$; $n=100\to2{,}7048\ldots$; $n=10\,000\to2{,}71814\ldots$
Notice how the sequence gets closer and closer to $e$ without ever exceeding it, however large you make $n$:
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:
$e_n = \left( 1 - \frac{1}{n} \right)^n$
The general formula for $1^{\infty}$
The same idea repeats itself with many different bases and exponents. There is a formula that turns any $1^{\infty}$ into a simpler limit, using exponent times (base minus 1).
Determine the limit of the following sequence, given 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, given that it presents an indeterminate form of the type $1^\infty$:
$f_n = \left( 1 + \frac{4}{n} \right)^n$
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 $1^\infty$:
$b_n = \left( \frac{2n+5}{2n+1} \right)^{3n-2}$
A more complicated algebraic exponent
Sometimes the exponent is neither $n$ nor a simple multiple of it, but an algebraic fraction. The procedure is identical: calculate $b_n\cdot(a_n-1)$ and solve that limit with the techniques you already know (degrees, conjugate…).
Find the limit of the sequence combining second-degree polynomials in the base: $c_n = \left( \frac{n^2+2n}{n^2-1} \right)^{5n}$
Find the limit of the sequence combining second-degree polynomials in the base:
$c_n = \left( \frac{n^2+2n}{n^2-1} \right)^{5n}$
Calculate the limit of the following sequence, which features an algebraic exponent in fractional form: $d_n = \left( \frac{n-2}{n+4} \right)^{\frac{n^2+1}{n}}$
Calculate the limit of the following sequence, which features an algebraic exponent in fractional form:
$d_n = \left( \frac{n-2}{n+4} \right)^{\frac{n^2+1}{n}}$
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:
$a_n = \left( \frac{2n^2-3n+1}{2n^2+n-5} \right)^{\frac{n^2+3}{n}}$
The number $e$ in continuously compounded interest
If you compound a capital $n$ times a year at a nominal interest rate $r$, the capital after one year is $C\cdot\left(1+\dfrac{r}{n}\right)^n$. When $n\to\infty$ (continuous compounding, instant by instant), the number $e$ appears.
An amount of $5000$ euros is deposited in a financial institution at an annual interest rate of $4\%$. The formula for the final capital after $n$ years with continuous…
An amount of $5000$ euros is deposited in a financial institution at an annual interest rate of $4\%$. The formula for the final capital after $n$ years with continuous compounding is given by the sequence $C_n = 5000 \cdot e^{0,04n}$. Calculate the accumulated capital after $10$ years and compare it with the capital obtained if the compounding were annual, whose expression is $C'_n = 5000 \cdot (1,04)^n$.
An amount of $10000$ euros is invested at a nominal annual interest rate of $5\%$. The sequence describing the final capital if interest is compounded $n$ times a year…
An amount of $10000$ euros is invested at a nominal annual interest rate of $5\%$. The sequence describing the final capital if interest is compounded $n$ times a year is $C_n = 10000 \left( 1 + \frac{0,05}{n} \right)^n$. Calculate the limit of this sequence as $n \to \infty$ to find the final capital under continuous compounding. Compare this result with the capital obtained if compounded quarterly ($n=4$).
Exponential growth and decay with $e$
Many natural processes (radioactive decay, cooling, pharmacology) are described directly by an exponential with base $e$: $a_n = a_0\cdot e^{kn}$, with $k>0$ for growth and $k<0$ for decay.
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? What value does the mass tend to in the long term (as $n \to \infty$)?
In a thermodynamics experiment, the temperature difference between a metallic body and its surroundings decreases exponentially according to the sequence…
In a thermodynamics experiment, the temperature difference between a metallic body and its surroundings decreases exponentially according to the sequence $T_n = 80 \cdot e^{-0,05n}$, where $n$ is the minutes elapsed since the start of cooling. Calculate the temperature difference after $30$ minutes and determine the limit of the sequence as $n \to \infty$.
Back to the beginning: If you invest at an annual interest rate $r$, paying the interest continuously does not send your profits up to infinity; instead, your final capital is bounded exactly by $C \cdot e^r$ (for example, at $5\,\%$, the return goes from $5\,\%$ a year to a ceiling of $5{,}13\,\%$).