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Next: 7. Still More Area Up: 6. Limits of Sequences Previous: 6.6 Geometric Series   Index

6.7 Calculation of $e$

6.81   Example. We will calculate $\displaystyle { \lim \left \{ n \ln \left( 1 + \frac{c}{n} \right) \right\} }$, where $c$ is a positive number. Let $\displaystyle { f(x) = {1 \over x} }$. Then (see the figure)

\begin{displaymath}B(1,1+{c \over n}: 0, {n \over n+c}) \subset S_1^{1+\frac{c}{n}}f
\subset B(1,1+ {c \over n}: 0,1)\end{displaymath}

and hence

\begin{displaymath}\mbox{\rm area}(B(1,1+{c \over n}: 0, {n \over n+c})) \leq \m...
\leq \mbox{\rm area}(B(1,1+ {c \over n}: 0,1)).


\begin{displaymath}\frac{c}{n} \cdot \frac{n}{n+c} \leq \ln(1 + \frac{c}{n})
\leq \frac{c}{n},\end{displaymath}

\begin{displaymath}\frac{cn}{n+c} \leq n\ln(1 + \frac{c}{n}) \leq c.
\end{displaymath} (6.82)


\begin{displaymath}\lim \left\{ {cn \over n+c} \right\} =
\lim \left\{ {c \over 1+{c\over n}} \right\} = c,\end{displaymath}

it follows from the squeezing rule that
\lim \left\{ n \ln\left( 1 + \frac{c}{n}\right)\right\} = c.
\end{displaymath} (6.83)

Notice that in this example the squeezing rule has allowed us to prove the existence of a limit whose existence was not obvious.

6.84   Example. We will show that for all $c \in \mbox{${\mbox{{\bf Q}}}^{+}$}$
\begin{displaymath}\lim \left\{ \left( 1 + \frac{c}{n}\right)^n \right\}
= e^c.
\end{displaymath} (6.85)

Let $c \in \mbox{${\mbox{{\bf Q}}}^{+}$}$, and let $n\in \mbox{{\bf Z}}^+$. Let

\begin{displaymath}a_n = \left( 1 + {c\over n}\right)^n.\end{displaymath}

By (6.82), we have

\ln(a_n) \leq c = c\ln(e) = \ln(e^c),


\begin{displaymath}a_n\leq e^c \mbox{ for all }n \in \mbox{{\bf Z}}^+.\end{displaymath}

It follows from (6.83) that
\lim\{ \ln(a_n)\} = c, \mbox{ or }\lim\{c-\ln(a_n)\} = 0.
\end{displaymath} (6.86)

From the picture, we see that

\begin{displaymath}0 \leq B(a_n,e^c:0, e^{-c}) \leq A_{a_n}^{e^c}\left[{1\over t}\right],\end{displaymath}


\begin{displaymath}0 \leq e^{-c}(e^c-a_n) \leq \ln(e^c) - \ln(a_n) = c -\ln(a_n).\end{displaymath}


\begin{displaymath}0 \leq e^c - a_n \leq e^c\big(c - \ln(a_n)\big), \mbox{ for all }n\in\mbox{{\bf Z}}^+.\end{displaymath}

By (6.86), we have

\begin{displaymath}\lim \{ e^c\big(c-\ln(a_n)\big)\} = 0,\end{displaymath}

so by the squeezing rule, $\lim\{ e^c-a_n\} = 0$, i.e.

\begin{displaymath}\lim\{a_n\} = e^c.\end{displaymath}

This completes the proof of (6.85).

The reason we assumed $c$ to be positive in the previous example was to guarantee that $(1+{c\over n})$ has a logarithm. We could extend this proof to work for arbitrary $c\in \mbox{{\bf Q}}^+$, but we suggest an alternate proof for negative $c$ in exercise 6.97

6.87   Example (Numerical calculation of $e$) It follows from the last example that

\begin{displaymath}\lim\left \{ \left( 1 + \frac{1}{n}\right)^n\right\} = e.\end{displaymath}

I wrote a Maple procedure to calculate $e$ by using this fact. The procedure limcalc(n) below calculates

\begin{displaymath}\left( 1 + \frac{1}{{100}^n} \right)^{{100}^n},\end{displaymath}

and I have printed out the results for n = 1,2,...,6.

> limcalc := n -> (1+ .01^n)^(100^n);

{\it limcalc} := {n} \rightarrow ( 1 + .01^{{n}} )^{( 100^{{n}
} )}

> limcalc(1);


> limcalc(2);


> limcalc(3);


> limcalc(4);


> limcalc(5);


> limcalc(6);


6.88   Exercise. From my computer calculations it appears that

\begin{displaymath}\lim\left \{ \left( 1 + \frac{1}{n}\right)^n\right\} = 1.\end{displaymath}

Explain what has gone wrong. What can I conclude about the value of $e$ from my program?

6.89   Example. Actually, Maple is smart enough to find the limit, and does so with the commands below. The command evalf returns the decimal approximation of its argument.

> limit( (1+1/n)^n,n=infinity);

{\rm e}

> evalf(%);


6.90   Entertainment ( $\lim \{n^{1\over n} \}$.) Find the limit of the sequence $\{n^{1\over n} \}$, or else show that the sequence diverges.

6.91   Example (Compound interest.) The previous exercise has the following interpretation.

Suppose that $A$ dollars is invested at $r$% annual interest, compounded $n$ times a year. The value of the investment at any time $t$ is calculated as follows:

Let $T = (1/n) \mbox{ year}$, and let $A_n^k$ be the value of the investment at time $kT.$ Then

$\displaystyle A_n^0$ $\textstyle =$ $\displaystyle A$  
$\displaystyle A_n^1$ $\textstyle =$ $\displaystyle A_n^0 + \frac{r}{100 n} A_n^0 = (1 + \frac{r}{100 n} ) A$  
$\displaystyle A_n^2$ $\textstyle =$ $\displaystyle A_n^1 + \frac{r}{100 n} A_n^1 = (1 + \frac{r}{100 n} )^2 A$ (6.92)

and in general
A_n^k = A_n^{k-1} + \frac{r}{100 n} A_n^{k-1} = (1 +\frac{r}{100 n} )^k A.\end{displaymath} (6.93)

The value of the investment does not change during the time interval $kT < t$
$ < (k+1)T.$ For example, if $V_n$ denotes the value of one dollar invested for one year at $r$% annual rate of interest with the interest compounded $n$ times a year, then

V_n = A_n^n = \left(1 + \frac{r}{100 n}\right)^n.

Thus it follows from our calculation that if one dollar is invested for one year at $r$% annual rate of interest, with the interest compounded ``infinitely often'' or ``continuously'', then the value of the investment at the end of the year will be

\begin{displaymath}\lim \left\{ \left(1 + {r\over 100n}\right)^n\right\} = e^{r\over 100}
\mbox{ dollars}.\end{displaymath}

If the rate of interest is 100%, then the value of the investment is $e$ dollars, and the investor should expect to get $2.71 from the bank.

This example was considered by Jacob Bernoulli in 1685. Bernoulli was able to show that $\displaystyle {\lim \left\{
\Big( 1+{1\over n}
\Big)^n \right\} < 3}$.[8, pp94-97]

6.94   Exercise. Calculate the following limits.
$\lim \{ {(1 + {3\over n})}^{2n} \}$.
$\lim \{ {(1 + {1\over 3n})}^{2n} \}$.

6.95   Exercise.
Use the formula for a finite geometric series,

\begin{displaymath}1 + (1-a) + (1-a)^2 + \cdots +(1-a)^{n-1} = { 1 - (1-a)^n \over 1-(1-a)}\end{displaymath}

to show that
(1-a)^n \geq 1 -na \mbox{ whenever $0 < a < 1$}.
\end{displaymath} (6.96)

Let $c \in \mbox{${\mbox{{\bf R}}}^{+}$}$ Use inequality (6.96) to show that

\begin{displaymath}\left(1 - {c\over n^2}\right)^n \geq 1 -{c\over n}\end{displaymath}

for all $n\in\mbox{${\mbox{{\bf Z}}}^{+}$}$ such that $n > \sqrt{c}$.
Prove that $\lim\{ (1-{c\over n^2})^n \} = 1 $ for all $c \in \mbox{${\mbox{{\bf R}}}^{+}$}$.

6.97   Exercise. A Let $c \in \mbox{${\mbox{{\bf Q}}}^{+}$}$. Use exercise 6.95 to show that

\begin{displaymath}\lim\left\{ {\left(1 - {c\over n}\right)}^n\right\} = e^{-c}.\end{displaymath}

(Hence we have $\lim \{ {(1+{c\over n})}^n \} = e^c \mbox{ for all }c \in \mbox{{\bf Q}}$.)

Hint: Note that $(1-z) = ({1-z^2\over 1+z})$ for all real numbers $z\neq -1$.

next up previous index
Next: 7. Still More Area Up: 6. Limits of Sequences Previous: 6.6 Geometric Series   Index
Ray Mayer 2007-09-07