Showing that $\lim_{n \rightarrow \infty} \left( 1 + \frac{r}{n} \right)^{n} = e^{r} $.

Question

I know that $ \displaystyle \lim_{n \rightarrow \infty} \left( 1 + \frac{1}{n} \right)^{n} = e $, but how do I show that $$ \lim_{n \rightarrow \infty} \left( 1 + \frac{r}{n} \right)^{n} = e^{r}? $$ Thanks!

Answer 1

$$ \lim_{n\to\infty}\left(1+\frac rn\right)^n=\lim_{n\to\infty}\left(\left(1+\frac rn\right)^{n/r}\right)^r =\left(\lim_{n/r\to\infty}\left(1+\frac rn\right)^{n/r}\right)^r=e^r $$

Answer 2

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.

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We start with $e$ being defined as the limit as $n\rightarrow \infty$ 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.

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

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

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,

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

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

$$ (1+1/n)^{nk} = (1+k/n)^n\left[1+O\left(1/n^2\right)\right]^n = (1+k/n)^n \left[1+O\left(1/n\right)\right] $$

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

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

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

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 $(1+p/qn)^n\rightarrow e^{p/q}$ as $n\rightarrow \infty$ which motivates us to take the $q$'th power of this expression in order to compare it with $e^p$.

$$ (1+p/nq)^{nq} = \left( 1+ p/n + \sum_{i=2}^q {q\choose i } \frac{p^i}{(nq)^i}\right)^n = \left(1+p/n + O\left( 1/n^2 \right)\right)^n = (1+p/n)^n\left(1+O\left(1/n^2\right)\right)^n = (1+p/n)^n \left(1+O\left(1/n\right) \right)$$

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 $(1-r/n)^n\rightarrow 1/e^r$.

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

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 $e^r= \lim_{n\rightarrow \infty} (1+r/n)^n$. 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,

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

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 $n\rightarrow \infty$ 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-r\vert < \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 $n\rightarrow \infty$ 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,

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

Answer 3

if $r>0$

Let $n=mr$, then $m\to\infty $ as $n\to\infty $ $$ \lim_{n\to\infty}\left(1+\frac rn\right)^n= \lim_{m\to\infty}\left(1+\frac {r}{rm}\right)^{rm}= \left(\lim_{m\to\infty}\left(1+\frac 1m\right)^m\right)^r=e^r $$

or In general

$$ \lim_{n\to\infty}\left(1+\frac rn\right)^n=\lim_{n\to\infty}e^{\ln\left(1+\frac rn\right)^n}=\lim_{n\to\infty}e^{\frac{\ln\left(1+\frac rn\right)}{\frac 1n}}=e^{\lim_{n\to\infty} \frac{\ln\left(1+\frac rn\right)}{\frac 1n}}=e^r $$ by using L'Hôpital's rule

Answer 4

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.

Answer 5

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

It is a try indeed, so be benevolent :p

By binomial formula $\displaystyle{\left(1+\frac xn\right)^n}=\sum\limits_{k=0}^{n} \mathsf{C_n^k}\frac{x^k}{n^k}=\sum\limits_{k=0}^{n} \frac{n!\;x^k}{k!\;(n-k)!\;n^k}=\sum\limits_{k=0}^{n}\frac{x^k}{k!}\times\frac{n!}{(n-k)!\;n^k}$

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

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

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

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

And finally we have gathered the following elements :

• $\displaystyle{\left(1+\frac xn\right)^n=\sum\limits_{k=0}^{n}a_{n,k}\frac{x^k}{k!}=\sum\limits_{k=0}^{\infty}a_{n,k}\frac{x^k}{k!}}$.

• $\displaystyle{|\sum\limits_{k=0}^{\infty}a_{n,k}\frac{x^k}{k!}|\le\sum\limits_{k=0}^{\infty}\frac{x^k}{k!}}=e^x\quad$ domination by a convergent serie.

• $\lim\limits_{n\to\infty}a_{n,k}=1\quad$ 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$\frac{n!}{n^n}\sim\sqrt{2\pi n}\;e^{-n}\to 0$), 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 $k\ll n$.

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

$\displaystyle{\lim\limits_{n\to\infty}\left(1+\frac xn\right)^n}=\sum\limits_{k=0}^{\infty}\lim\limits_{n\to\infty}\left[a_{n,k}\frac{x^k}{k!}\right]=\sum\limits_{k=0}^{\infty}\frac{x^k}{k!}=e^x$.

Answer 6

The question has been answered for positive $r$. A small modification takes care of negative $r$. So we want to find $$\lim_{n\to\infty}\left(1-\frac{w}{n}\right)^n,$$ where $w\gt 0$. Whenever $n\ne w$, 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 $n\to\infty$. 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.