My professor gave us the following proof problem:
Prove that the sequence $a_n=\left ( 1+\frac{1}{n}\right)^n$ converges to $e$.
Instruction:
Prove that the sequence is monotonic increasing and bounded above.
Now I understand how to prove that it's increasing and bounded above, but how does proving that entail that the sequence converges to $e$?
IF we assume that we know what $e$ is and we're not treating this problem as it's unknown and we're trying to calculate it's numerical value, you can verify it converges to $e$ treating it as a simple limit, as:
$$\lim_{n\to\infty }\left(1+\frac{1}{n}\right)^n=\lim_{n\to\infty }e^{n\ln\left(1+\frac{1}{n}\right)}=\lim_{n\to\infty }e^\frac{\ln\left(1+\frac{1}{n}\right)}{\frac{1}{n}}=\lim_{x\to 0}e^{\frac{\ln(1+x)}{x}},$$
where we let $x=\dfrac{1}{n}$.
Then, it is
$$\lim_{x\to 0} \frac{\ln(1+x)}{x} = \lim_{x \to 0} \frac{1}{1+x} = 1,$$
by L'Hospital's rule.
So, all in all:
$$\lim_{x\to 0}e^{\frac{\ln(1+x)}{x}}=e^1=e.$$
On the other hand, if you want a "precise" elaboration on calculating the numerical value via series, the user mathlove has posted a great answer on this post, which I am just pasting below for the sake of completeness of my attempted answer:
\begin{align}\left(1+\frac 1n\right)^n&=\sum_{k=0}^{n}\binom{n}{k}\left(\frac 1n\right)^k\\&=\sum_{k=0}^{n}\frac{n(n-1)\cdots(n-k+1)}{k!}\left(\frac 1n\right)^k\\&=\sum_{k=0}^{n}\frac{1}{k!}\left(1-\frac 1n\right)\left(1-\frac 2n\right)\cdots\left(1-\frac{k-1}{n}\right)\\&\lt \sum_{k=0}^{n}\frac{1}{k!}\\&=1+\frac{1}{1}+\frac{1}{2}+\frac{1}{3\cdot 2}+\frac{1}{4\cdot 3\cdot 2}+\cdots+\frac{1}{n\cdot(n-1)\cdots 2}\\&\lt 1+\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots+\frac{1}{2^{n-1}}\\&=3-\frac{1}{2^{n-1}}\\&\lt 3.\end{align}
After all, it really comes down to what the user Randall said in the comments.
$$\begin{align*} \lim_{n\to\infty}a_n &= \lim_{n\to\infty}\left(1 + \frac1n\right)^n \\ \text{ Binomial Expansion }\\ \lim_{n\to\infty}\left(1 + \frac nn + \frac{n(n-1)}{2!n^2} + \frac {n(n-1)(n-2)}{3!n^3}+...\right) &\implies \lim_{n\to\infty}(1 + \frac11 + \frac {1^2}{2!} + \frac {1^3}{3!} + ...) = \color{blue}{2.7182818284....}\end{align*}$$
I don't know how you conceive this thing but this is how I had done when I hadn't studied calculus and series
Let us consider that the above series will keep giving me a different number but then I realized that the factorial increases drastically high dividing poor small number 1. (maybe after that they named it e: )[E for Euler]
$$\lim\sum F = \sum\lim F \text{ *same thing }$$
The monotone convergence theorem states the following: if $a_n\lt{U}$ for some $U\in\mathbb{R}$ and $a_n\lt{a_{n+1}}$ for every $n\in\mathbb{N}$, then $a_n$ converges to some $L\in\mathbb{R}$. By the monotone convergence theorem, the fact that $a_n$ is bounded from above and is monotonic implies that it converges to some real number. $e$ is defined as the real number this sequence converges to.
As for why the monotone convergence theorem is true, that is a separate question in and of itself. I am not going to provide a rigorous proof here, but instead, I want to show that it makes sense intuitively. If $a_n$ is monotonically increasing, then it is mostly well-behaved: it is not oscillating wildly or anything, and if it is bounded, then it is not exploding to $\infty$. Both results combined give you that $a_n\lt{a_{n+1}}\lt{U}$, and as $n\to\infty$, $a_n$ gets bigger, and thus, closer to $U$, thus $a_{n+1}$ is being sort of squished in between. Since it is being squished in between, it has no choice but to approach something, because $U-a_n$ is getting smaller. In particular, you can always choose your upper bound $U$ so that $U-a_n$ gets arbitrarily small.
