Two Roads to e

First, $2 < e < 3$: $e = \sum_{k=0}^{\infty}\frac{1}{k!} > 1+1 = 2$, and since $k! \ge 2^{k-1}$ for $k \ge 1$, $e < 1+1+\frac{1}{2}+\frac{1}{4}+\cdots = 3$. So if $e = \frac{p}{q}$ then $q \ge 2$.

$q!e = (q-1)!\,p$ is an integer, and so is $\sum_{k=0}^{q}\frac{q!}{k!}$. Put $T = q!\left(e - \sum_{k=0}^{q}\frac{1}{k!}\right)$; then $T = \frac{1}{q+1}+\frac{1}{(q+1)(q+2)}+\cdots$, so $0 < T < \frac{1}{q+1}\left(1+\frac{1}{q+1}+\cdots\right) = \frac{1}{q} \le \frac{1}{2} < 1$.

But $q!e = \text{(integer)} + T$ with $0 < T < 1$ is not an integer — a contradiction. Hence $e$ is irrational.

Assume $n \ge 1$ (for $n = 0$, $\frac{1}{n \cdot n!}$ is undefined and the geometric bound $\frac{n+1}{n}$ below is invalid). $R_n = \frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\cdots = \frac{1}{(n+1)!}\left(1+\frac{1}{n+2}+\frac{1}{(n+2)(n+3)}+\cdots\right)$.

The bracket is less than $1+\frac{1}{n+1}+\frac{1}{(n+1)^2}+\cdots = \frac{n+1}{n}$ (geometric series).

Hence $R_n < \frac{1}{(n+1)!}\cdot\frac{n+1}{n} = \frac{1}{n\cdot n!}$. This estimate is exactly why the factorial series wins by a landslide.

Put $s_n = \left(1+\frac{1}{n}\right)^n$. By the binomial theorem $s_n = \sum_{k=0}^{n}\binom{n}{k}\frac{1}{n^k}$, and since $\binom{n}{k}\frac{1}{n^k} = \frac{1}{k!}\prod_{j=0}^{k-1}\left(1-\frac{j}{n}\right)$ (empty product $= 1$ for $k = 0$), $s_n = \sum_{k=0}^{n} a_{n,k}$ with $a_{n,k} = \frac{1}{k!}\prod_{j=0}^{k-1}\left(1-\frac{j}{n}\right)$.

Termwise comparison: for fixed $k \ge 1$ and $0 < j \le k-1$, $\left(1-\frac{j}{n}\right) < \left(1-\frac{j}{n+1}\right)$ (larger $n$ subtracts less), so $a_{n,k} \le a_{n+1,k}$ — and for $k \ge 2$ the factor $j = 1$ is strictly smaller, giving $a_{n,k} < a_{n+1,k}$. All terms are positive.

The exact difference is $s_{n+1} - s_n = \sum_{k=0}^{n} (a_{n+1,k} - a_{n,k}) + a_{n+1,n+1}$.

Every difference $a_{n+1,k} - a_{n,k} \ge 0$ and the extra term $a_{n+1,n+1} > 0$, so $s_{n+1} - s_n > 0$.

For $n \ge 2$ the $k = 2$ difference $a_{n+1,2} - a_{n,2} > 0$ also contributes positively.

The case $n = 1$ is checked directly: $s_1 = 1 + 1 = 2$ and $s_2 = 1 + 1 + \frac{1}{4} = 2.25$, so $s_2 - s_1 = \frac{1}{4} > 0$ (the term $a_{1,2}$ does not exist in the $n = 1$ sum, so applying the $k = 2$ gain at $n = 1$ would double-count $a_{2,2}$).

So $s_n$ is strictly increasing.

Each factor satisfies $\left(1-\frac{j}{n}\right) \le 1$.

For $k = 0$ the product is empty, so $a_{n,0} = \frac{1}{0!} = 1$; for $k = 1$ the only factor is $j = 0$, giving $1$, so $a_{n,1} = \frac{1}{1!} = 1$ — equality holds.

For $k \ge 2$ the factor $j = 1$ is strictly below $1$, so $a_{n,k} < \frac{1}{k!}$.

Hence $a_{n,k} \le \frac{1}{k!}$ (equality only for $k = 0, 1$), and $s_n \le \sum_{k=0}^{n}\frac{1}{k!} < \sum_{k=0}^{\infty}\frac{1}{k!} = e$ — $s_n$ stays below $e$, and its limit $L = \lim s_n$ satisfies $L \le e$.

It remains to show the limit is exactly $e$.

Fix $K$; for $n \ge K$, $s_n \ge \sum_{k=0}^{K} a_{n,k}$.

For fixed $k$, as $n \to \infty$ each factor $\left(1-\frac{j}{n}\right) \to 1$, so $a_{n,k} \to \frac{1}{k!}$.

Hence $L \ge \sum_{k=0}^{K} \frac{1}{k!}$ for every $K$, and letting $K \to \infty$ gives $L \ge e$.

Combined with $L \le e$ from the previous step, $L = e$ — $s_n$ increases to $e$ from below.