Jtdylktuy

$$
lim_{ntoinfty}left(1+frac rnright)^n=lim_{ntoinfty}left(left(1+frac rnright)^{n/r}right)^r
=left(lim_{n/rtoinfty}left(1+frac rnright)^{n/r}right)^r=e^r
$$

This is my take on the issue. I'm assuming the existence of all the limits involved simply because I just don't want to type that much more.

I am not using the exponential function to define exponents instead I am taking the definitions I used in this answer. Basically positive integer powers are repeated multiplication, rational powers through $n'th$ roots, real powers by least upper bounds. Negative powers are taken to be the reciprocals of the positive powers.

I may have made some typos or errors in judgement since the answer is quite long. I am happy to fix any issues which are brought to my attention.

We start with $e$ being defined as the limit as $nrightarrow infty$ of $(1+1/n)^n$. The goal is to establish that $(1+r/n)^n rightarrow e^r$ for rational $r$ and to then use this result to define real powers of $e$. As a starting point we consider integer powers of $e$ first.

Positive Integer Powers
Suppose that $k$ is a positive integer and consider the $k$'th power of $e$ defined as the product of $e$ with itself $k$ times.

$$ e^k = prod_{i=1}^k left[ lim_{nrightarrow infty}(1+1/n)^n right] = lim_{nrightarrow infty} prod_{i=1}^k left[ (1+1/n)^n right] = lim_{nrightarrow infty} left[ prod_{i=1}^k (1+1/n) right]^n $$

Note that,
$$prod_{i=1}^k (1+1/n) = (1+1/n)^k = sum_{p=0}^k {kchoose p}frac{1}{n^p} = 1+frac{k}{n}+O(1/n^2) = (1+k/n)left[1+Oleft(frac{1}{n^2(1+k/n)}right)right] $$

Note that $ 1/n^2 geq 1/[n^2(1+k/n)]$ which means we can replace the $O(1/[n^2(1+k/n)])$ with $O(1/n^2)$. So that we have,

$$(1+1/n)^k = (1+k/n)left[1+Oleft(1/n^2right)right]$$

Taking the $n$'th power of this expression we get,

$$ (1+1/n)^{nk} = (1+k/n)^nleft[1+Oleft(1/n^2right)right]^n = (1+k/n)^n left[1+Oleft(1/nright)right] $$

$$ Rightarrow vert (1+1/n)^{nk} - (1+k/n)^n vert = O(1/n) $$

This tells us that the limit of $(1+1/n)^{nk}$ is the same as the limit of $(1+k/n)^n$ as $nrightarrow infty$. Therefore we conclude that,

$$boxed{ e^k = lim_{nrightarrow infty} (1+1/n)^{nk} = lim_{nrightarrow infty} (1+k/n)^n} $$

Positive Rational Powers

To include positive rational powers we just need to establish that $e^{p/q}$ is the $q$'th root of $e^p$. We suspect that $(1+p/qn)^nrightarrow e^{p/q}$ as $nrightarrow infty$ which motivates us to take the $q$'th power of this expression in order to compare it with $e^p$.

$$ (1+p/nq)^{nq} = left( 1+ p/n + sum_{i=2}^q {qchoose i } frac{p^i}{(nq)^i}right)^n = left(1+p/n + Oleft( 1/n^2 right)right)^n = (1+p/n)^nleft(1+Oleft(1/n^2right)right)^n = (1+p/n)^n left(1+Oleft(1/nright) right)$$

We can then use this to show that $ vert (1+p/nq)^{nq} - (1+p/n)^n vert = O(1/n)$ which by the same argument as above tells us that the two expressions have the same limit.

So we now have that the limit of $(1+p/nq)^n$, which we will call $L$, has the property that $L^q=e^p$. Therefore $L$ is the $q$'th root of $e^p$ and we indicate this with the notation $e^{p/q}$.

$$ boxed{(1+p/nq)^n rightarrow e^{p/q}} $$
** Negative Rational Powers **

Negative rational powers of a number are defined as the reciprocals of the positive rational powers of the same number. Therefore given a positive rational number $r$ we wish to show that $(1-r/n)^nrightarrow 1/e^r$.

$$ left( lim_{nrightarrow infty} (1-r/n)^n right) left( lim_{nrightarrow infty} (1+r/n)^n right) = lim_{nrightarrow infty} left[ (1-r/n) (1+r/n)right]^n = lim_{nrightarrow infty} left[ 1-r^2/n^2right]^n = lim_{nrightarrow infty} left[ 1-O(1/n) right] = 1 $$

Therefore the limit of $(1-r/n)^n$ is the reciprocal of $e^r$ and we identify it with the notation $e^{-r}$.

$$ boxed{(1-r/n)^n rightarrow e^{-r}}$$

** Real / Irrational Powers of e **

We now understand that for rational powers $e^r= lim_{nrightarrow infty} (1+r/n)^n$. We wish to show that this still holds for irrational powers $x$.

The definition of a real power $e^x$ is in terms of the least upper bound property of the real numbers,

$$ e^x = sup lbrace e^r mid rin Q, r < x rbrace $$

Our job is then to show that the limit of $(1+x/n)^n = L$ is the least upper bound of the set indicated above.

Note that given a rational number $r$ which is less than $x$ we have, $(1+r/n) < (1+x/n)$ which tells us that $(1+r/n)^n < (1+x/n)^n$. Allowing $nrightarrow infty$ we get the inequality $e^r leq L $. We have therefore established that $L$ is an upperbound on the indicated set of rational powers of $e$.

To show that $L$ is the least upper bound we must show that no number less than $L$ can be an upper bound of the set. We will establish this by showing that $e^r$ can be made arbitrarily close to $L$ by choosing an $r$ close to $x$.

Suppose that $r<x$ and $vert x-rvert < delta $ then we have,

$$ vert (1+x/n)^n - (1+r/n)^n vert = vert 1+x/n-1-r/n vert vert sum_{i=1}^{n} left[(1+x/n)^{n-i}(1+r/n)^{i-1} right] vert $$

$$ leq frac{vert x-r vert}{n} sum_{i=1}^{n} vert[(1+x/n)^{n-1} vert leq frac{delta}{n} n vert (1+x/n)^{n-1} vert = delta (1+x/n)^{n-1}$$

Allowing $nrightarrow infty$ we get,

$$vert e^r - L vert leq delta L$$

Which means that $e^r$ can be put within $epsilon >0$ of $L$ if we choose $r$ to be within $delta = epsilon L$ of $x$. Since epsilon can be made arbitrarily small we know that for any $alpha<L$ we can find a $alpha < e^r <L$. That is to say that $L$ is a least upper bound or the set of rational powers $e^r$ with $r<x$. Since this is the definition of $e^x$ we have shown that,

$$ boxed{ (1+x/n)^n rightarrow e^x } $$

if $r>0$

Let $n=mr$, then $mtoinfty $ as $ntoinfty $
$$
lim_{ntoinfty}left(1+frac rnright)^n= lim_{mtoinfty}left(1+frac {r}{rm}right)^{rm}= left(lim_{mtoinfty}left(1+frac 1mright)^mright)^r=e^r
$$

or In general

$$ lim_{ntoinfty}left(1+frac rnright)^n=lim_{ntoinfty}e^{lnleft(1+frac rnright)^n}=lim_{ntoinfty}e^{frac{lnleft(1+frac rnright)}{frac 1n}}=e^{lim_{ntoinfty} frac{lnleft(1+frac rnright)}{frac 1n}}=e^r
$$
by using L'Hôpital's rule

The question has been answered for positive $r$. A small modification takes care of negative $r$. So we want to find
$$lim_{ntoinfty}left(1-frac{w}{n}right)^n,$$
where $wgt 0$. Whenever $nne w$, we have
$$left(1-frac{w}{n}right)^n=frac{1}{left(1+frac{w}{n-w}right)^n}.$$
We want to find the limit of the denominator as $ntoinfty$. Note that
$$left(1+frac{w}{n-w}right)^n =left(left(1+frac{w}{n-w}right)^{n-w}right)^{n/(n-w)}.$$
Now the result follows from the solutions for positive $r$ already posted.

It is best to assume that $r$ is a real number and from the context the variable $n$ is an integer. But many answer have not taken the restriction of $n$ as an integer into account.

Also it looks much better if we replace $r$ by $x$. Then we consider the limit $$F(x) = lim_{n to infty}left(1 + frac{x}{n}right)^{n}$$ We know that $F(0) = 1$ and $F(1) = e$ (this can be taken as definition of $e$).

The next step is show that $F(x)$ is defined for all $x$ (apart from $x = 0, 1$). This is bit tricky but can be done using the fact that $F(x, n) = (1 + (x/n))^{n}$ is increasing and bounded if $x > 0$ and hence it tends to a limit $F(x)$ as $n to infty$.

For negative $x = -y$, it makes sense to study the function $G(y, n) = (1 - (y/n))^{-n}$ which is decreasing and bounded below and so tends to a limit $L$. Since $G(y, n) = 1/F(x, n)$, it follows that $F(x, n)$ tends to limit $1/L$.

So $F(x)$ exists for all real $x$. Next we can show $F(x + y) = F(x)F(y)$. Using this we can easily show that if $x$ is rational then $F(x) = {F(1)}^{x} = e^{x}$. For irrational $x$ we define the irrational power $e^{x}$ by the expression $F(x)$. The proof that $F(x + y) = F(x)F(y)$ is again tricky and is available here. The development of these ideas is done in reasonable detail in my blog post.

Let's try to do it via the power serie definition : $displaystyle{e^x=sumlimits_{k=0}^{infty}frac{x^k}{k!}}$.

It is a try indeed, so be benevolent :p

By binomial formula $displaystyle{left(1+frac xnright)^n}=sumlimits_{k=0}^{n} mathsf{C_n^k}frac{x^k}{n^k}=sumlimits_{k=0}^{n} frac{n!;x^k}{k!;(n-k)!;n^k}=sumlimits_{k=0}^{n}frac{x^k}{k!}timesfrac{n!}{(n-k)!;n^k}$

Let's call :
$begin{cases}
a_{n,k}=frac{n!}{(n-k)!;n^k} quad mathrm{for}quad kle n \
a_{n,k}=0 qquadqquad mathrm{for}quad k>n \
end{cases}$

This is $n(n-1)(n-2)...(n-k+1)/n^k$

So written like this it is easy to see that for each $iin{0,1,..,k-1}$ we have $0le(n-i)le n$ and consequently $0le a_{n,k}le 1$.

When $k$ is fixed and $kll n$ (which is realized when $ntoinfty$) we have also $a_{n,k}to 1$.

And finally we have gathered the following elements :

• $displaystyle{left(1+frac xnright)^n=sumlimits_{k=0}^{n}a_{n,k}frac{x^k}{k!}=sumlimits_{k=0}^{infty}a_{n,k}frac{x^k}{k!}}$.




• $displaystyle{|sumlimits_{k=0}^{infty}a_{n,k}frac{x^k}{k!}|lesumlimits_{k=0}^{infty}frac{x^k}{k!}}=e^xquad$ domination by a convergent serie.




• $limlimits_{ntoinfty}a_{n,k}=1quad$ simple convergence of serie terms

Here we have a perfect setup to apply dominated convergence theorem.

This theorem is more powerful and thus more convenient than trying to use a uniform convergence argument on $a_{n,k}$.

One may claim that when $k$ approaches $n$ then $a_{n,k}to 0$ and not $1$ (since$frac{n!}{n^n}simsqrt{2pi n};e^{-n}to 0$), but $k$ is only an artefact of the discrete measure used for series instead of proper integrals, thus all is going as if we were taking $lim_n a_{n,k}$ with a fixed $k$, keeping us in the favorable domain where $kll n$.

So applying the theorem we get that the limit of LHS actually exists and that we can swap limit and summation.

$displaystyle{limlimits_{ntoinfty}left(1+frac xnright)^n}=sumlimits_{k=0}^{infty}limlimits_{ntoinfty}left[a_{n,k}frac{x^k}{k!}right]=sumlimits_{k=0}^{infty}frac{x^k}{k!}=e^x$.