Proof of $\lim_{n \to \infty} \left( 1+ \frac{r}{n}\right)^n = e^r$ for $r \in \mathbb Q$ from "basic" principles.

Question

I am trying to prove the fact that $\lim_{n \to \infty} \left( 1+ \frac{r}{n}\right)^n = e^r$ for $r \in \mathbb Q$ using only the fact that $\lim_{n\to \infty} (1+ 1/n)^n = e$ and "basic" facts of limits, such that for $r \in \mathbb Q$ and $(x_n)\to x$, we have $(x_n)^r = x^r$. I appreciate that there are at least two posts about this (e. g. here). I find the following proof particularly nice:

Setting $m = n/r$ gives $$ \lim _{n \rightarrow \infty}\left(1+\frac{r}{n}\right)^n= \lim _{n \rightarrow \infty}\left(1+\frac{1}{n / r}\right)^n= \lim _{n \rightarrow \infty}\left(1+\frac{1}{m}\right)^{m r}= \lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{m}\right)^m\right]^r=e^r $$

There are two steps in this proof that seem non trivial to me though (and are very closely linked):

1. Why does the fact that $\lim_{n\to \infty} (1 + 1/n)^n$ imply that $\lim_{x \to \infty}(1 + 1/x)^x$ if we take $f : x \mapsto (1+1/x)^x$ to a function defined on $\mathbb R$?
2. Why may we assume that just because $m(n) = n/r$ grows as $n$ does, that the limit stays the same when we substitute $n$ for $m$?

Answer 1

The way it should be done is as following :

1. Prove that $$\lim_{n\to \infty}\left(1+\frac{k} {n}\right)^n=e^k,$$ for all $k\in\mathbb Z$.

2. Prove that $$\lim_{n\to \infty }\left(1+\frac{1/p}{n} \right)^n=e^{\frac{1}{p}},$$ for all $p\in\mathbb N$.

3. Prove that $$\lim_{n\to \infty }\left(1+\frac{r}{n}\right)^n=e^r,$$ for all $r\in \mathbb Q$.

Answer 2

Edit: I restructured the argument to make it clearer.

There are four basic facts about that we need here. Let $(a_n)_n$ be a sequence of real numbers, then:

1. If there exists a positive integer $k$ and a real number $L$ such that for all $0\leq i<k$ the subsequence $(a_{nk+i})_n$ converges to $L$, then $(a_n)_n$ converges to $L$ as well
2. If $(a_n)_n$ is convergent, then for any subsequence converges to the same limit.
3. If $(i_n)_n$ and $(s_n)_n$ are convergent sequences with same limit $L$ and if $i_n\leq a_n\leq s_n$ for all $n$, then also $(a_n)_n$ is convergent with limit $L$.
4. If $I\subseteq\mathbb{R}$ is a closed interval, $f\colon I\to\mathbb{R}$ is a continuous function and $(a_n)_n$ converges to some real number $L$, then $(f(a_n))_n$ is convergent with limit $f(L)$.

Now to the proof. For simplicty, I assume $r\in\mathbb{Q}_{>0}$, so that $r=p/q$ with $p,q\in\mathbb{Z}_{>0}$. Denote $a_n=\left(1+\frac{r}{n}\right)^n$. To show convergence, we use Fact 1 above. To cancel the denominator $p$ of $r$, we consider subsequences of the form $(a_{pn+i})_n$ with $0\leq i<p$ fixed. Note that $$ a_{pn+i}=\left(1+\frac{p/q}{pn+i}\right)^{pn+i}=\left(1+\frac{1}{q(n+i/p)}\right)^{pn+i} $$ and thus $$ \left(1+\frac{1}{q(n+1)}\right)^{p(n+1)+i-p}\leq a_{pn+i}\leq\left(1+\frac{1}{qn}\right)^{pn+i}. $$ Hence if we define $c_n=\left(1+\frac{1}{qn}\right)^{pn}$, then we have $$ \left(1+\frac{1}{q(n+1)}\right)^{i-p}c_{n+1}\leq a_{pn+i}\leq \left(1+\frac{1}{qn}\right)^{i}c_n. $$ Now we want to use Fact 3. Notice that $\lim_{n\to\infty}\left(1+\frac{1}{q(n+1)}\right)^{i-p}=1$ and $\lim_{n\to\infty}\left(1+\frac{1}{qn}\right)^{i}=1$, so if we can show that $(c_n)_n$ converges to some limit $L$, both the lower and upper bound converge to $L$, and Fact 3 will give that also $(a_{pn+i})_n$ converges to $L$.

To obtain convergence of $(c_n)_n$, we introduce $b_n=\left(1+\frac{1}{n}\right)^n$ and use the fact that $(b_n)_n$ converges to $e$. To make a $qn$ appear, we consider the subsequence $b_{qn}$ and use Fact 2: we have $$ \lim_{n\to\infty}\underbrace{\left(1+\frac{1}{qn}\right)^{qn}}_{=b_{qn}}=\lim_{n\to\infty}\underbrace{\left(1+\frac{1}{n}\right)^{n}}_{=b_n}=e. $$ Notice that $b_{qn}$ almost looks like $c_n$, except that the exponent isn't quite right. To fix this, we consider the function $f\colon[0,+\infty)\to\mathbb{R}$ defined by $f(x)=x^r$. Notice that $$ f(b_{qn})=f\left(\left(1+\frac{1}{qn}\right)^{qn}\right)=\left(1+\frac{1}{qn}\right)^{rqn}=\left(1+\frac{1}{qn}\right)^{pn}=c_n. $$ As $f$ is continuous, we obtain by Fact 4 that $(c_n)_n$ is convergent and $$ \lim_{n\to\infty}\underbrace{c_n}_{=f(b_{qn})}=f\left(\lim_{n\to\infty}b_{qn}\right)=f(e)=e^r. $$ Therefore, by Fact 3 and our estimate in the beginning, we obtain also $$ \lim_{n\to\infty}a_{pn+i}=e^r $$ for every $0\leq i<p$. Therefore, by Fact 1, we conclude $$ \lim_{n\to\infty}\underbrace{\left(1+\frac{r}{n}\right)^n}_{a_n}=e^r. $$ To obtain the result also for negative $r$, notice that we have $$ \left(1-\frac{1}{n}\right)^{n}=\left(\frac{n-1}{n}\right)^{n}=\frac{1}{\left(\frac{n}{n-1}\right)^{n}}=\frac{1}{\left(1+\frac{1}{n-1}\right)^{n}}. $$ With similar techinques, this let's you deduce that $\left(1-\frac{1}{n}\right)^{n}$ converges to $e^{-1}$, and then you can run the same argument as above for negative $r$.

Answer 3

To both your questions: Yes, these steps need more justification. Neither follows from the "first-principles definition" that $\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n=e$. A typical justification would go as follows:

For $x \geq 1$, we have $$\begin{align*}\left(1+\frac1{\lceil x\rceil}\right)^{-1} \left(1+\frac1{\lceil x\rceil}\right)^{\lceil x \rceil} &\leq \left(1+\frac1{\lceil x\rceil}\right)^{\lfloor x \rfloor} \\ &\leq \left(1+\frac1x\right)^{\lfloor x \rfloor} \\ &\leq \left(1+\frac1x\right)^x \\ &\leq \left(1 + \frac1x\right)^{\lceil x\rceil} \\ &\leq \left(1 + \frac1{\lfloor x\rfloor}\right)^{\lceil x\rceil} \\ &\leq \left(1+\frac1{\lfloor x\rfloor}\right)\left(1 + \frac1{\lfloor x\rfloor}\right)^{\lfloor x\rfloor} \end{align*}$$

Now, as $x\to\infty$, both $\lfloor x\rfloor$ and $\lceil x\rceil$ tend to $\infty$ through the integers, so in particular, the two outer-most expressions in the previous chain of inequalities tend to $1\cdot e = e$. By the squeeze/sandwich theorem, we may conclude $$\lim_{x\to\infty} \left(1+\frac1x\right)^x = e.$$

As an immediate consequence, we can now use the sequence of equalities you were asking about to show $$\lim_{x\to\infty}\left(1+\frac{r}{x}\right)^x = e^r$$ for any real $r>0$ (or $r\geq 0$ once you notice $r=0$ is trivial).

This does leave $r<0$ to be shown, but hopefully that was enough to satisfy your curiosity.