Limit of sequences: $ \lim (1 + \frac{r}{n} )^n = e^r $

Question

Consider the sequence $x_n = \left (1+\frac{r}{n} \right )^n $ for $r \in \mathbb{Q} $. I need to prove that $ \lim x_n = e^r $

My attempt of proof (for r>0) is to find a subsequence of $x_n$ that converges to $e^r$, but it will be enough if $x_n$ is a Cauchy sequence. Since all the convergent sequences are Cauchy sequences, I was thinking to prove that $x_n$ is convergent and it will finish my problem. So if I prove that that $x_n$ is increasing and have an upper bound ($3^r$, for example), then it is done. But I'm having trouble with this last step.

But maybe it is easier to prove using, for example, that $\left(1+\frac{r}{n}\right)^n = \left(1+\frac{1}{n/r}\right)^n$ but I don't see how to proceed too.

I appreciate your help.

Thanks!

Answer 1

I will use the following definitions. $\log$ is defined as $$ \log(x) = \int_1^x \frac{1}{t} \, dt. $$ for all $x > 0$. From this definition it can be shown that all the familiar properties of $\log$ holds. $e^x$ is defined as $e^x = \log^{-1}(x)$ for all $x \in \mathbb{R}$ (it can be proved that $\log$ is increasing so its one to one and maps onto $\mathbb{R}$, so the inverse exists). Then $e$ is defined as $e = e^1 = \log^{-1}(1)$.

Now I will prove the limit. By the fundamental theorem of calculus, $\log'(\frac{1}{r}) = r$. Using the definition of derivatives, we have $$ \lim_{h \to 0} \frac{\log(\frac{1}{r} + h) - \log(\frac{1}{r})}{h} = r \\ \lim_{h \to 0} \frac{1}{h}\log(1 + rh) = r \\ \lim_{h \to 0} \log\left((1 + rh)^{\frac{1}{h}}\right) = r. $$ Since $\log$ is continuous, we can write $$ \log\left(\lim_{h \to 0} (1 + rh)^{\frac{1}{h}}\right) = r \\ \lim_{h \to 0} (1 + rh)^{\frac{1}{h}} = e^r. $$ So we have $\lim_{n \to \infty} (1 + \frac{r}{n})^n = e^r \\$ where the expression inside the limit is a function. So it must also be true for the sequence with integer $n$.

---

EDIT:

Let $a_n = \left(1 + \frac{r}{n}\right)^n$. This sequence is increasing. To see this, note that $(1 + a)^b \geq 1 + ab$ (by binomial theorem). Apply this for $a = \frac{r}{n + 1}$ and $b = \frac{n + 1}{n}$ to get $$ \left(1 + \frac{r}{n + 1}\right)^{\frac{n + 1}{n}} \geq 1 + \frac{r}{n + 1} \cdot \frac{n + 1}{n} = 1 + \frac{r}{n}. $$ Taking both sides to the power of $n$ gives $\left(1 + \frac{r}{n + 1}\right)^{n + 1} \geq \left(1 + \frac{r}{n}\right)^n$ which shows that the sequence is increasing. To show that it is bounded, first use the binomial theorem to get $$ \left(1 + \frac{r}{n}\right)^n = \sum_{i = 0}^n \binom{n}{i} \left(\frac{r}{n}\right)^i = \sum_{i = 0}^n \frac{n!}{i!(n - i)!} \cdot \frac{r^i}{n^i} \\ = \sum_{i = 0}^n \frac{r^i}{i!} \left(\frac{n(n - 1) \dotsb (n - i + 1)}{n^i}\right) \leq \sum_{i = 0}^n \frac{r^i}{i!}. $$ since $n(n - 1) \dotsb (n - i + 1) \leq \underbrace{n \dotsb n}_{i} = n^i$. So $$ \left(1 + \frac{r}{n}\right)^n \leq \sum_{i = 0}^n \frac{r^i}{i!} < \sum_{i = 0}^{\infty} \frac{r^i}{i!}. $$ But the series on the right most side converges which we can check by the ratio test. So $a_n$ is bounded by that number. Thus $a_n$ converges since it is increasing and bounded.

Answer 2

If you define $e = \lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^n$, then you have (for $r\neq 0$) $$\lim_{n\rightarrow\infty}\left(1+\frac{r}{n}\right)^n = \lim_{n\rightarrow\infty}\left(1+\frac{1}{\frac{n}{r}}\right)^{\frac{n}{r}r} = \lim_{n\rightarrow\infty}\left(1+\frac{1}{k}\right)^{kr} = e^r$$ where we have written $k=\frac{n}{r}$.

Answer 3

If you use the definition $e =: \lim_{n\to \infty} \left(1+ \frac{1}{n} \right)^n $for $r \in \Bbb Q $ and $n \in \Bbb N $, choose your sub-sequence $m \in \Bbb N $ such that $n/r = m$ is a natural number the take $m \to \infty$ to get $e^r$ like in Daniel Robert-Nicoud's answer.

EDIT:: It is given that $a_n = \left( 1 + \frac 1 n \right)^n$ converges to $e$. let $m \in \Bbb N $ be a subsequence of $n \in \Bbb N$ such that $m/r = n$ a natural number. $x_n = \left( 1 + \frac r n \right)^n$, then take sub-sequence $x_m$ of $x_n$, you get $x_m = \left( 1 + \frac 1 n \right)^m = \left( 1 + \frac 1 n \right)^{nr } = (a_n)^r$. We know that $a_n$ converges to $e$. Now it remains to show that, $\{(a_n)\}^r$ converges to $e^r$.

It is also given that $r$ is a rational number, so let $r = \frac{p}{q} \;, p, q \in \Bbb N$. \begin{align*} \left | a_n^r - e^r\right | &= \left|(a_n^{1/q})^p - (e^{1/q})^p\right| \\ & \le \left|(a_n^{1/q}) - (e^{1/q})\right|\cdot |p e^{(p-1)/q}|\\ & \le |a_n - e| \cdot \frac{q e^{q-1}}{\sum_{k=0}^{q-1} e^{k/q}} \\ &< \epsilon \\ \end{align*} for some $n \ge N \in \Bbb N$.