0

In my problem book, there was a question: By defining $e= \lim\limits_{n \to \infty}\left( 1+\frac{1}{n} \right) ^n$ prove that $e^x = \sum\limits_{n=0}^ \infty \frac{x^n}{n!}$. this is a strange question because this book is all about sequence of real numbers and derivatives or Taylor's theorem aren't introduced yet

Here is my attempt:

Lemma 1:- The sequence $e_n := \left( 1+\frac{1}{n} \right) ^n$ converges.

Proof: By binomial theorem $$e_n =1+1 +\frac{1-\frac{1}{n}}{2!} +\frac{(1-\frac{1}{n})(1-\frac{2}{n})}{3!} +\frac{(1-\frac{1}{n})(1-\frac{2}{n})(1-\frac{3}{n})}{4!}+ \dots $$ ,clearly $e_{n+1}> e_n$ then the sequence is strictly increasing. $e_n \leq 2+\sum_{k=2}^ \infty \frac{1}{k!}$ it is obvious that $n!>2^{n-1}$ then $e_n<2+\sum_{k=1}^ \infty \frac{1}{2^k} =3$ by monotone convergence theorem $e_n $ converge to $e:= \sup\{{e_n} \}$

Lemma 2:- the sequence $a_n:= \left( 1-\frac{1}{n} \right)^n $ converges.

Proof: note that $a_n<1$ so the sequence is bounded above, $$\frac{a_{n+1}}{a_n}=\left( \frac{n}{n-1} \right)^\frac{1}{n} \left( \frac{n}{n+1} \right)^{\frac{1}{n+1}} = \left(\left( \frac{n}{n-1} \right)^{n+1} \left( \frac{n}{n+1} \right)^{n} \right)^{\frac{1}{(n)(n+1)}}$$ $$ =\left( \frac{1}{(1-\frac{1}{n^2})(1-\frac{1}{n})}\right)^{\frac{1}{(n)(n+1)}} \geq 1$$ then the sequence is strictly increasing by monotone convergence theorem $a_n $ converge to $a:= \sup\{{a_n} \}$.

Corollary:- the sequence $a_n:=\left( 1-\frac{1}{n} \right)^n$ converges to $\frac{1}{e}$.

Proof: $$a_n=\left(\frac{n-1}{n} \right)^{{n}}$$ Let $k=n+1$ $$a_n=\left(\frac{k}{k+1} \right)^{{k+1}}=\left(\frac{k+1}{k} \right)^{-k-1} $$ $$\lim_{n\to \infty} a_n = \frac{1}{e}$$

Lemma 3 For any monotone increasing sequence $x_n\to \infty$ ,then $\lim\limits_{n\to \infty} \left(1+ \frac{1}{x_n} \right)^{{x_n}}= e$

Proof:- Define $\alpha_n= \lfloor x_n\rfloor $ , $\beta_n=\lceil x_n\rceil $

$$\left(1+ \frac{1}{\alpha_n} \right)^{\alpha_n -1}\leq\left(1+ \frac{1}{x_n} \right)^{{x_n}}\leq \left(1+ \frac{1}{\beta_n} \right)^{\beta_n +1}$$

since $\left(1+ \frac{1}{\alpha_n} \right)^{\alpha_n-1} ,\left(1+ \frac{1}{\beta_n} \right)^{\beta_n +1} $ is a sub sequence of $\left(1+ \frac{1}{n} \right)^{n-1} , \left(1+ \frac{1}{n} \right)^{n+1}$ respectively
then by squeeze theorem $\lim\limits_{n\to \infty} \left(1+ \frac{1}{x_n} \right)^{{x_n}}= e$

Corollary:- $e^x = \lim\limits_{n\to \infty} \left(1+ \frac{x}{n} \right)^{n}$

Proof By lemma 3 $$e^x = \left(\lim\limits_{n\to \infty} \left(1+ \frac{x}{n} \right)^{\frac{n}{x}}\right)^{x}=\lim\limits_{n\to \infty} \left(\left(1+ \frac{x}{n} \right)^{\frac{n}{x}}\right)^{x}=\lim\limits_{n\to \infty} \left(1+ \frac{x}{n} \right)^{n}$$

Here I couldn't rigorously prove that $ \lim\limits_{n\to \infty} \left(1+ \frac{x}{n} \right)^{n}=\sum\limits_{n=0}^ \infty \frac{x^n}{n!}$

pie
  • 4,192
  • How is the exponentiation defined in your textbook? – Sangchul Lee Dec 19 '23 at 07:33
  • @SangchulLee $$e= \lim\limits_{n \to \infty}\left( 1+\frac{1}{n} \right) ^n$$ $$e^x= \lim\limits_{n \to \infty}\left( 1+\frac{1}{n} \right) ^{xn}$$ – pie Dec 19 '23 at 07:38
  • 1
    I mean, how is the operation $a^x$ defined in your textbook when $x$ is not an integer? Or is your textbook taking this operation (and some of its basic properties) taken granted without establishing them? We can't prove something without knowing how that is defined. – Sangchul Lee Dec 19 '23 at 07:39
  • Why would you manipulate the limit definition of $e^x$ to get something that is not $e^x$? It should be $\lim_{n \rightarrow \infty } (1 + \frac{x}{n})^{\frac{n}{x}}$ no? – Robert Murray Dec 19 '23 at 07:41
  • 1
    Various proofs here: https://math.stackexchange.com/q/69806/42969, and here: https://math.stackexchange.com/q/2348229/42969, and here: https://math.stackexchange.com/q/754673/42969. – Martin R Dec 19 '23 at 07:41
  • @SangchulLee I think they are taken granted without establishing them, btw I don't know how to define $a^b$. can you recommend me some reference. – pie Dec 19 '23 at 07:41
  • 1
    There are two main (pedagogical) ways of introducing exponentiation. In the first approach, one subsequently enlarges the range of $x$ for which $a^x$, $a>0$, is defined, starting from when $x\in\mathbb{N}0$. In the final step of this construction, assuming $a^x$ has been successively defined for rational $x$, the case of real $x$ can be defined by $$a^x=\lim{\mathbb{Q}\ni r\to x}a^r,$$ where the limit is taken along any sequence of rational numbers $r$ that converges to $x$. It can be shown that this limit converges and satisfies all the desired properties of exponentiation. – Sangchul Lee Dec 19 '23 at 07:50
  • 1
    But, given that you are allowed to use any familiar properties of exponentiation to your liking, I think the answers linked in Martin R's comment will an answer your question. (In hindsight, I think what I really wanted to know was what familiar properties of exponentiation are allowed to be used without proving them, rather than how exponentiation is defined in your context.) – Sangchul Lee Dec 19 '23 at 07:51

1 Answers1

1

Two hints to help you: One, you may have trouble proving $$\lim_{n \rightarrow \infty} (1+ \frac{x}{n})^{\frac{1}{n}} = \lim_{n \rightarrow \infty} \sum_{k=0}^n \frac{x^k}{k!}$$As the left hand side is not $e^x$. In the last corrolary you made some unseen mistake and changed $e^x$s definition (if we want $\frac{x}{n} =\frac{1}{x_n} $, we need $x_n = \frac{n}{x}$).

Second, When you have come to the correct definition of $e^x$, we may find: $$ (1+ \frac{x}{n})^n = 1 + \frac{ {n \choose 1} x}{n} + \ldots + \frac{{ n \choose k} x^k}{n^k}$$ Via the binomial expansion. And for large $n$, $n^k$ actually dissapears when combined the $n \choose k$, which you can read about here

Robert Murray
  • 1,788