I'm very confused about the question below, which I couldn't figure out for days.
In Example 5 the author is teaching us proving
$$\lim_{n\to \infty} (1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...\frac{1}{n!}) = e$$
with equation
$$x_n = (1+\frac{1}{n})^n$$ $$y_n = 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...\frac{1}{n!}$$
I'm thinking $x_n < e$ if we take $e$ is the special value only $n$ tends to infinity, and term by term comparing tells us $x_n \le y_n$ if we consider $n$ can be $1$, then we have the following: $x_n \le y_n, x_n < e$, but I believe that doesn't give us $e\le y_n$. I think only if we use $x_n = e$ and with $x_n \le y_n$ brings us $e \le y_n$. So I'm quite confused how did him resulted $e \le e$ first?
Then $m < n$ is implemented to help us go further, with $x_m < x_n < e$ and resulting $y_m\le e$? How? I'm thinking such outcomes only come from $x_m \le y_m$ and $x_m < x_n \le e$.
I really need your help.
Let $x_n=(1+\frac{1}{n})^n.$By the binomial theorem, we have
$x_{n}=\sum_{k=0}^n\binom{n}{k}\frac{1}{n^k}=\sum_{k=0}^n\frac{1\cdot(1-\frac{1}{n})\cdots(1-\frac{k-1}{n})}{k!}$
Find an integer $m$ such that $m<n$,there is
$x_n>\sum_{k=0}^m\frac{1\cdot(1-\frac{1}{n})\cdots(1-\frac{k-1}{n})}{k!}$
Let $n\rightarrow \infty$,then $\mathrm{e}\geq \sum_{k=0}^m\frac{1}{k!}$.Similarly, let $m\rightarrow \infty$,we can get $\mathrm{e}\geq \sum_{n=0}^\infty\frac{1}{n!}$
On the other hand,
$x_{n}=\sum_{k=0}^n\frac{1\cdot(1-\frac{1}{n})\cdots(1-\frac{k-1}{n})}{k!}\leq\sum_{k=0}^n\frac{1}{k!}$
Let $n\rightarrow \infty$ ,we get $\mathrm{e}\leq \sum_{n=0}^\infty\frac{1}{n!}$.Hence , by the sandwich theorem,we have
$\mathrm{e}=\sum_{n=0}^\infty\frac{1}{n!}$