Validity and Equivalence of two definitions of the real exponential function

Question

The Problem :

We state the following two definitions of the real exponential function from the Pr$\infty$fWiki page. We're interested in showing that the two definitions are valid $($i.e. the defining sequence/series does converge to a unique real number$)$ and that the two definitions are equivalent.

I'm stuck at a couple of points $($which are described in highlighted lines$)$. Any help would be much appreciated. Thank you!

Definition $1$. The exponential function can be defined as the following limit of a sequence $$\exp x := \lim_{n \to \infty} \left({1 + \frac x n}\right)^n$$

Definition $2$. The exponential function can be defined as a power series $$\exp x := \sum_{n = 0}^\infty \frac {x^n} {n!}$$

My Progress and two places where I'm stuck : Essentially the solution consists of three parts, namely validity of definition $1$, validity of definition $2$ and equivalence of the two definitions.

Validity of Definition $1$. I'm stuck here! Can we show that the sequence $(a_n)$ given by $a_n = \left(1+\frac{x}{n}\right)^n$ converges for every $x \in \mathbb{R} ??$ Is it eventually monotone and bounded $??$

Validity of Definition $2$. The radius of convergence of the power series is $$r=\lim_{n \to \infty}\left| \frac{\frac{1}{n!}}{\frac{1}{(n+1)!}} \right|=\lim_{n \to \infty}\left| \frac{(n+1)!}{n!} \right|=\lim_{n \to \infty}(n+1)=+\infty$$ Thus the infinite series in the right-hand side of Definition $2$ converges to a unique real number for all $x \in \mathbb{R}$. Hence the definition is well-defined.

Equivalence of Definition $1$ and Definition $2$.

There's this proof of Definition $1$ $\implies$ Definition $2$ in the Pr$\infty$fWiki page, but it's kind of under construction and I'm not really convinced by it. So I decided to try to take my own shot at it.

For all $n \in \mathbb{N} \cup \{0\},$ let $$T_n=\left(1+\frac{x}{n} \right)^n, ~S_n=\sum_{k=0}^n \frac{x^k}{k!}$$ We have to show that $\lim_{n \to \infty} T_n = \lim_{n \to \infty} S_n$

Now, \begin{align} T_n &= \left(1+\frac{x}{n}\right)^n\\ &= 1+n\cdot\frac{x}{n}+\frac{n(n-1)}{2!}\cdot\frac{x^2}{n^2}+\cdots +\frac{n(n-1)\cdots 1}{n!}\cdot\frac{x^n}{n^n}\\ &= 1+x+\left(1-\frac{1}{n}\right)\cdot \frac{x^2}{2!}+\cdots +\left(1-\frac{1}{n}\right)\cdots \left(1-\frac{n-1}{n}\right)\cdot \frac{x^n}{n!} \end{align}

Clearly, $$S_n-T_n=\left\{1-\left(1-\frac{1}{n}\right)\right\}\frac{x^2}{2!}+\cdots +\left\{1-\left(1-\frac{1}{n}\right)\cdots \left(1-\frac{n-1}{n}\right)\right\}\cdot \frac{x^n}{n!}\geq 0$$

I'm stuck at this point. Can we show that $S_n-T_n \leq B_n$ such that $B_n \to 0$ as $n \to \infty ??$

Answer 1

Let $E(x)=\sum_{j=0}^{\infty}x^j/j!.$

Given any $x\in \mathbb R$ and given any $\epsilon >0,$ take $k_1\geq 1$ such that $$(I).\quad \sum_{j=1+k_1}^{\infty }|x|^j/j!<\epsilon.$$

For brevity let $1-\prod_{i=0}^{j-1}(1-i/n)= A(n,j)$ for $1\leq j\leq n.$

Take $k_2>k_1$ such that $$(II). \quad n\geq k_2\implies \max \{\;|(1-A(n,j)\cdot x^j/j!|\;: 1\leq j\leq k_1\}<\epsilon /k_1.$$

For $n\in \mathbb N$ we have $$E(x)-(1+x/n)^n=\sum_{j=0}^{\infty}B_jx^j$$ where $B_0=0$ and $B_j=(1-A(n,j))/j!$ for $1\leq j\leq n$ and $B_j=1/j!$ for $j>n.$

Observe that $0\leq B_j\leq 1/j!$ for all $j.$

Now when $n\geq k_2$ we have $$|\sum_{j=0}^{k_1}B_jx^j|= |\sum_{j=1}^{k_1}B_jx^j|\leq \sum_{j=1}^{k_1}|B_jx^j|=$$ $$=\sum_{j=1}^{k_1}|(1-A(n,j)\cdot x^j/j!|<\epsilon \quad \text {by }\; (II),$$ and also we have $$|\sum_{j=1+k_1}^{\infty}B_jx^j |\leq \sum_{j=1+k_1}^{\infty}|B_jx^j|\leq$$ $$ \leq \sum_{j=1+k_1}^{\infty}|x^j/j!|<\epsilon \quad \text { by }\; (I).$$ So $|E(x)-(1+x/n)^n|<2\epsilon$ for all $n\geq k_2.$

Answer 2

Starting point: $E(x) = \sum_{n = 0}^\infty \frac {x^n} {n!}$ converges.

1st Question: Show that $a_n = \left(1+\frac{x}{n}\right)^n$ converges.

Solution: Show that $(a_n)$ converges to $E(x)$.

Look at Konrad Königsberger's analysis, just tone it down a bit by replacing the $w$ stuff with $x$.

Lemma: ${n \choose k} \frac 1 {n^k} = \frac{1}{k!} \prod_{i = 1}^{k-1} \Bigl(1 - \frac i n \Bigr) \le \frac 1 {k!}$
Moreover, the sequence $t_n = {n \choose k} \frac 1 {n^k}$ converges to $\frac {1}{k{!}}$ as $n \to +\infty$.

The OP should be familiar with the triangle inequality for infinite sums.

If $K$ and $n$ are positive integers and $n \ge K$ we have a RHS side broken down into three sums:

$\left|\Bigl(1 + \frac{x}{n}\Big)^n - E(x) \right| \le \sum_{k=0}^{K-1} \left|{n \choose k}\frac{x^k}{n^k} - \frac{x^k}{k!}\right| + \sum_{k=K}^n{n\choose k} \frac{|x|^k}{n^k} + \sum_{k=K}^\infty \frac{|x|^k}{k!}$

So far this is just expressive notation showing the binomial expansion of $a_n$ combined with $E(x)$ (in a useful way) while using the triangle inequality.

Let $\varepsilon \gt 0$ be given. We choose $K$ so that the last term is less than $\varepsilon/3$. By the Lemma, the middle term will also be less than $\varepsilon/3$ (true for any selected $n$). By the Lemma, we can choose $N$ so that for all $n \gt N$ the first term is also less than $\varepsilon/3$.