Proof of $\lim_{n \to \infty}(1+\frac{1}{n})^n=e$ I have this messy proof of $\lim_{n \to \infty}(1+\frac{1}{n})^n=e$ in my notebook. I can't find it anywhere else, but I need it since the professor accepts only this version at the exam. At the time I am only stuck with the first part, so I will write the proof only to that point. Also, there is a few steps with missing reasoning between them.
Proof.
We are given two sequences:
$$x_n=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots+\frac{1}{n!}$$
$$y_n=\left(1+\frac{1}{n}\right)^n$$
If $x_n$ converges, then $\lim_{n \to \infty}x_n=e$. (This is given.)
Let's prove that $x_n$ converges.
To prove that, the Monotone Convergence Theorem will be used.
Since $x_{n+1}>x_{n}$ the sequence is increasing. (I skipped the unnecessary reasoning for this.)
Now let's prove that it is bounded from above.
$$n!=1\cdot 2\cdot 3\cdots n>2^{n-1}\tag{1.1}$$
$$\frac{1}{n!}\leq \frac{1}{2^{n-1}}\tag{1.2}$$
$$\underbrace{2}_{\text{?}}\leq x_n\leq 1+1+\frac{1}{2}+\frac{1}{2^2}+\dots+\frac{1}{2^{n-1}}\tag{1.3}$$
$$1+q+q^2+\dots+q^{n-1}=\underbrace{\overbrace{\frac{1-q^n}{1-q}}^{?}=\frac{q^n-1}{q-1}}_{\text{?}}\tag{1.4}$$
$$q=\frac{1}{2}$$
$$\underbrace{1}_{\text{?}}+\frac{1-\frac{1}{q^n}}{1-\frac{1}{q}}=1+2\left(1-\frac{1}{2^n}\right)<1+2=3\tag{1.5}$$
And so $2\leq x_n\leq3$, $x_n$ is bounded and increasing, hence it converges. Therefore $\lim_{n \to \infty}x_n=e$.
The proof proceeds with proving that $\lim_{n\to \infty}y_n=Y$, and then showing $y\geq e$ and $y \leq e$, and thereby $y=e$. This part is very long and already looking messy in the notebook. Because of that I will skip it, since it isn't important for the question.

My questions:

*

*Is there any obvious reason why $2^{n-1}$ is a good (necessary?) choice for the inequality (1.1)?


*How does (1.2) follow from (1.1)? Why is it $\leq$ and not $ < $ ? I see that $\leq$ is the right choice since for $n=1$, we have $\frac{1}{1!}=\frac{1}{2^{1-1}}$, but is there any way to know this without evaluating for choices of $n$?


*Why is that $2$ there in inequality (1.3)? Where did it come from ?


*Considering inequality (1.4) there is few things;
a) How was the overbraced fraction derived from the geometric series?
b) The underbraced equality at first didn't make sense to me, but when I evaluated for few $n$s, I realized it actually holds. Is this a known thing, that the order of subracting isn't important for two fractions to be equal, as long as both the numerator and the denominator of the fraction are both $<0$ or $>0$? Also what was the role of the equality in the proof, or was it just a remark ?
EDIT: I just realized that we can do that since the fraction is always positive. But it still doesn't seem so obvious to notice. Should it be obvious?


*Why is that $1$ there in the inequality (1.5) ? Where did it come from ?


*If you have a source for this proof please send me a link or tell me where I can find it.
Thanks
 A: *

*$2^{n-1}$ is a good choice because of the series in (1.3)


*I think that it is a typo. The correct is $n!\geq 2^{n-1}$, since for $n=1$, the equality holds. And for all the values, we know that $n!>2^{n-1}$. It is a classical exercise in Analysis (maybe Calculus too?) that the function $x!$ grows faster than $x^a$ for any real value of $a$. Try it :)


*Remember that $x_n=1+\dfrac{1}{1!}+\cdots+\dfrac{1}{n!}$. For $n=1, x_1=2$, and for all other values for $n$, $x_n>2$. That's way $x_n\geq 2.$


*(a) In general,
$$a+ar+\cdots+ar^{n-1}=a\dfrac{1-r^n}{1-r}. $$
There is several ways of proving this. For example, let $S_n=a+ar+\dots ar^n.$ Now evaluate $S_n-S_{n-1}$, and see what you got.
(b) Just multiply the numerator and the denominator by $(-1)$


*It comes from "nowhere". It is the beginning of a new equation. It just uses the inequality in (1.4)


*I don't know anyone :c
A: I'll try answer step by step:

*

*$(1.1)$ is not true for $n=1$.


*To obtain $(1.2)$ from $(1.1)$ use $\frac{1}{a} <\frac{1}{b}\Leftrightarrow b<a$ for non negative numbers.


*For left side of $(1.3)$ we have that $x_1$ already contain 2, for right use $(1.2)$


*There is formula for sum for first members of geometric progression


*It is first member of $x_n$


*Rudin W. - Principles of mathematical analysis-(1976) from page 63 have similar reasonings.
