Show that ##\lim_{n\to\infty}(1+\frac{r}{n})^n=e^r##

[McFluffy]

Homework Statement

The constant $e$ can be defined in many ways. My first exposure to this number involved compound interests. Specifically, if you decide to continuously compound 1$ at a 100% interest rate for a certain period of time(year, month, etc), you'll end up with $e$$ where it is defined as: $\lim_{n\to\infty}(1+\frac{100\%}{n})^{n}=e$.

I've been told that if you wanted to continuously compound some amount of money, $P$ at some interest rate per unit time, $r$ and that you want to do it $t$ units of time you'll get: $P\cdot e^{rt}$. I understand why there's an exponent $t$ there because of how multiplication is defined but what I don't understand is why is $r$ there. I'm trying to do this using only basic properties about limits and not other tools from calculus if possible as I'm not well adept at it yet. So the problem statement that I asked myself is:

How do you show that $e^{r} = \lim_{n\to\infty} (1+\frac{r}{n})^n$?

Homework Equations

$\lim_{n\to\infty}(1+\frac{1}{n})^n =e$
$\sum_{n=0}^{\infty} \frac{1}{n!} = e$
$(a+b)^{n} = \sum_{k=0}^{n}$ $n \choose {k}$ $a^{n-k}b^{k}$,

3. The Attempt at a Solution

I tried expanding $(1+\frac{r}{n})^n$ using the binomial theorem to show that it is $e^{r}$ from there using its other definition which $\sum_{n=0}^{\infty} \frac{1}{n!}$ but I all I got was useless manipulations of the equation with tedious writings and was pretty much lost on how to get to there.

 

[PSRB191921]

Hi
you can write $$\left( 1+\frac{r}{n} \right) ^{n}=exp\left[ n.ln\left( 1+\frac{r}{n} \right) \right] $$
and after that it is simple
PSR

 

[McFluffy]

Hi
you can write $$\left( 1+\frac{r}{n} \right) ^{n}=exp\left[ n.ln\left( 1+\frac{r}{n} \right) \right] $$
and after that it is simple
PSR

Thanks but I'm stuck; I tried simplifying $n\ln{(1+\frac{1}{n})}$ but I couldn't get it to equal to $r$.

 

[PSRB191921]

remember that:
$$\lim \nolimits_{n\to \infty }ln\left( 1+\frac{r}{n} \right) \to \frac{r}{n} $$
it helps ?

 

[McFluffy]

remember that:
$$\lim \nolimits_{n\to \infty }ln\left( 1+\frac{r}{n} \right) \to \frac{r}{n} $$
it helps ?

Thank you for that.

 

[Ray Vickson]

Homework Statement

The constant $e$ can be defined in many ways. My first exposure to this number involved compound interests. Specifically, if you decide to continuously compound 1$ at a 100% interest rate for a certain period of time(year, month, etc), you'll end up with $e$$ where it is defined as: $\lim_{n\to\infty}(1+\frac{100\%}{n})^{n}=e$.

I've been told that if you wanted to continuously compound some amount of money, $P$ at some interest rate per unit time, $r$ and that you want to do it $t$ units of time you'll get: $P\cdot e^{rt}$. I understand why there's an exponent $t$ there because of how multiplication is defined but what I don't understand is why is $r$ there. I'm trying to do this using only basic properties about limits and not other tools from calculus if possible as I'm not well adept at it yet. So the problem statement that I asked myself is:

How do you show that $e^{r} = \lim_{n\to\infty} (1+\frac{r}{n})^n$?

Homework Equations

$\lim_{n\to\infty}(1+\frac{1}{n})^n =e$
$\sum_{n=0}^{\infty} \frac{1}{n!} = e$
$(a+b)^{n} = \sum_{k=0}^{n}$ $n \choose {k}$ $a^{n-k}b^{k}$,

3. The Attempt at a Solution

I tried expanding $(1+\frac{r}{n})^n$ using the binomial theorem to show that it is $e^{r}$ from there using its other definition which $\sum_{n=0}^{\infty} \frac{1}{n!}$ but I all I got was useless manipulations of the equation with tedious writings and was pretty much lost on how to get to there.

Note: this is a significantly simpler (edited) version of the previous response under this same #.

If you can accept that $e = \lim_{x \to \infty} (1 + \frac{1}{x})^x$ holds for $x$ going to $\infty$ through non-integer values of $x$ (so that changing $n$ to $x$ has no effect), then you are part essentially done, at least for rational values of $r$.

First, if $r$ is a positive rational, then $e = \lim_n (1+r/n)^{n/r}$, by taking $x = n/r \to \infty$. For any finite $n>1$ we have $[(1+r/n)^{n/r}]^r = (1+r/n)^n,$ so $e^r = \lim_n (1+r/n)^n$.

Next, if $r > 0$ is rational, we have
$$ \left( 1 - \frac{r}{n} \right)^n \left( 1 + \frac{r}{n} \right)^n = \left( 1 - \frac{r^2}{n^2} \right) ^n, $$
and this last quantity $\to 1$ as $n \to \infty.$ (This holds because
$$ \left( 1-\frac{b}{n^2} \right)^n = \left[ \left( 1 - \frac{b}{n^2} \right)^{n^2} \right]^{1/n},$$ so if
$m = \lim (1 - b/n^2)^{n^2} > 0$ we have that $(1-b/n^2)^n$ is close to $m^{1/n}$, and this $\to 1$.)

Therefore, we have that $\lim_n (1-r/n)^n (1+r/n)^n =1$, so $\lim_n (1-r/n)^n = 1/e^r = e^{-r}$. Thus, the desired limit result holds as well for negative (rational) values of the exponent.

To prove that the result holds also for irrational values of $r$ involves some calculus arguments that go a bit beyond caclulus 101.

 

[McFluffy]

Note: this is a significantly simpler (edited) version of the previous response under this same #.First, if $r$ is a positive rational, then $e = \lim_n (1+r/n)^{n/r}$, by taking $x = n/r \to \infty$. For any finite $n>1$ we have $[(1+r/n)^{n/r}]^r = (1+r/n)^n,$ so $e^r = \lim_n (1+r/n)^n$.

.

Thanks for the lengthy answer. There some parts where I don't understand but my immediate question is that why did you set the condition that $n>1$ and not $n>0$? What's the reason behind it?

 

[Ray Vickson]

Thanks for the lengthy answer. There some parts where I don't understand but my immediate question is that why did you set the condition that $n>1$ and not $n>0$? What's the reason behind it?

Well, we are interested in $n \to \infty$, so small $n$ are of no interest. If you prefer, we could restrict $n$ to $n > 10,000,000$.

However, a more fundamental reason is that certain exponent laws are still true, but vacuous (i.e., contain no useful information) when we have an exponent of 0 For example, $x^0 = 1$ and $y^0 = 1$ for any $x, y > 0$, so an equation like $x^0 = y^0$ tells us nothing at all about how $x$ and $y$ are related.

Sometimes we need to treat $n=0$ separately. For example, we can define $n!$ as the product of all the integers from $1$ to $n$, but that does not tell use anything about whether $0!$ has meaning and what it should be. Usually, however, we define $0!=1$, and that is definitely not the product of the integers between 1 and 0. Using $0!=1$ allows us to say $n! = n (n-1)!$ seamlessly for $n = 1,2,3, \ldots$, and we often need to use something like that.

So: the moral of the story is that we sometimes need to treat $n=0$ differently from all other integers, and just developing the habit of doing that can save a lot of work and worry. For that reason I just automatically adopt the restriction $n \geq 1$ as a default whenever including $n = 0$ would serve no useful purpose.

 

[McFluffy]

What topics are required to learn in order to know how to prove $r$ holds for irrational numbers?