Proof of $\lim_{n \to \infty}(1+\frac{1}{n})^n=e$

Question 1) we want

$x_n = 1 + \frac 1{1!} + \frac 1{2!} + ..... + \frac 1{n!} \le $

$1 + f(1) + f(2) + ...... + f(n) \le$

$1+F(n)$

where $F(n)$ is something bounded that we can manipulate.

$f(k) = \frac 1{2^{k-1}}$ and $F(n) = 1+ \frac 11+ \frac 12 + \frac 14 + ..... + \frac 1{2^{n-}}$ are good choices because $F(n) = 1+ \frac 11+ \frac 12 + \frac 14 + ..... + \frac 1{2^{n-1}}= 3-\frac 1{2^{n}}<3$ is a really easy calculation to make and prove, and proving that $\frac 1{n!} =\frac 1{2*3*4*5*....*n} \le \frac 1{2*2*2*.....*}=\frac 1{2^{n-1}}$ is equally easy.

If we can find find some other $f(k)$ and $F(n)$ and $K$ so that $\frac 1{n!} \le f(n)$ and that $1 + \sum_{k=1}^n f(k) = 1+ F(n) \le K$ we are free to use those but... powers of $2$ these choices are so easy.

Question 2:

$\le$ is weaker than $<$ and a stronger statement always implies a weaker statement.

$n! > 2^{n-1} \implies \frac 1{n!} < \frac 1{2^{n-1}} \implies \frac 1{n!} \le \frac 1{2^{n-1}}$.

That certainly isn't false.

But as to why we used $\le$ rather than $<$..... well, for the case when $n =1$ and $\frac 1{n!} = \frac 1{2^{n-1}}$. That's all. It's a one time exception.

Question 3:

It's just a lower bound. If $n = 1$ then $x_1 = 1 + \sum_{k=1}^1 \frac 1{k!} = 1+ \frac 1{1!} = 2$.

That's all.

Question 4:

It is a well known equality that for any $q\ne 1$ that $1 + q + q^2 + ..... + q^{n-1} = \frac {1-q^n}{1-q} = \frac {q^n -1}{q-1}$

$(1+q + q^2 + ..... + q^{n-1})(1-q)=$

$(1+q + q^2 + ..... + q^{n-1}) - q(1+q + q^2 + ..... + q^{n-1})=$

$(1+q + q^2 + ..... + q^{n-1}) - (q + q^2 + q^3 + ..... + q^{n})= $

$(1 + \underbrace{q^2 + ..... + q^{n-1}}) -(\underbrace{q^2 + ..... + q^{n-1}} + q^n) =$

$1-q^n$.

So if $(1+q + q^2 + ..... + q^{n-1})(1-q) = 1-q^n$ then $\frac {1-q^n}{1-q} = (1+q + q^2 + ..... + q^{n-1})$

"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"

Yes. It is a known thing: $\frac {a-b}{c-d} = \frac {a-b}{c-d}\frac {-1}{-1} = \frac {-(a-b)}{-(c-d)} = \frac {b-a}{d-c}$.

Question 5:

If $\color{green}{1 + q + q^2 + ..... + q^{n-1}} = \color{green}{\frac {q^n -1}{q-1}}$ then

$\color{red}1 + \color{green}{1 + q + q^2 + ..... + q^{n-1}} = \color{red}1 +\color{green}{\frac {q^n -1}{q-1}}$

And notice back in 1.3 we had

$x_n = \color{red}1 + \color{green}{1 + \frac 12 + \frac 1{4} + .... + \frac 1{2^{n-1}}}$

....

Maybe this is all to complicated.

I'd do it.

$1 + \frac 1{1!} + \frac 1{2!} + \frac 1{3!} + \frac 1{4!}+ ....... + \frac 1{n!} =$

$(1 + 1) + \frac 1{2} + \frac 1{2*3} + \frac 1{2*3*4} + ..... + \frac 1{2*3*4*....n}< $

$2 + \frac 1{2} + \frac 1{2*2} + \frac 1{2*2*2} + .... + \frac 1{2*2*2*....*2} =$

$2 + (\frac 12 + \frac 1{4} + \frac 1{8} + ...... + \frac 1{2^{n-1}})$.

Then I'd make one of my flying leaps that every one knows the things I used to know as a kid that $\frac 12 + \frac 1{4} + \frac 1{8} + ...... + \frac 1{2^{n-1}} = 1-\frac 1{2^n} < 1$ because everyone has heard of the jumping flea paradox so everyone has seen this in kindergarten, haven't they? No? .... Well, each time we add only half of what we need to get to $1$ so we never add enough to get to one so the sum has to be less than $1$, right? Because we never add enough to get to $1$.

So $2 + (\frac 12 + \frac 1{4} + \frac 1{8} + ...... + \frac 1{2^{n-1}})< 2 + 1=3$

That really is all this proof is saying.

The key issue is noting that $2*3*4*...... *n \le 2*2*2*2*....*2$. Everything else just falls into place.