The series for e and the bound on the remainder

The exponential series

The starting point of the proof is the series expansion of $e$, obtained by evaluating the exponential series at $1$:

$$e = \sum_{k=0}^{+\infty} \frac{1}{k!} = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \frac{1}{120} + \cdots$$

Convergence is very fast. Let us denote the partial sums and the remainder:

$$S_n = \sum_{k=0}^{n} \frac{1}{k!} \qquad \text{and} \qquad R_n = e - S_n = \sum_{k=n+1}^{+\infty} \frac{1}{k!}$$

$n$	$S_n$	$e - S_n$
$2$	$\dfrac{5}{2} = 2.5$	$\approx 2.18 \times 10^{-1}$
$4$	$\dfrac{65}{24} \approx 2.70833$	$\approx 9.95 \times 10^{-3}$
$6$	$\dfrac{1957}{720} \approx 2.71806$	$\approx 2.26 \times 10^{-4}$
$10$	$\approx 2.7182818$	$\approx 2.73 \times 10^{-8}$

The decisive bound on the remainder

The whole proof rests on the following inequality, valid for $n \geq 1$:

$$0 < R_n < \frac{1}{n \cdot n!}$$

Lower bound. All the terms of $R_n$ are strictly positive, so $R_n > 0$.

Upper bound. Factor out the first term:

$$R_n = \frac{1}{(n+1)!}\left(1 + \frac{1}{n+2} + \frac{1}{(n+2)(n+3)} + \cdots\right)$$

Each factor $\dfrac{1}{n+j}$ is strictly less than $\dfrac{1}{n+1}$ as soon as $j \geq 2$, so the parenthesis is bounded above by the geometric series with ratio $\dfrac{1}{n+1}$:

$$R_n < \frac{1}{(n+1)!} \sum_{j=0}^{+\infty} \frac{1}{(n+1)^j} = \frac{1}{(n+1)!} \cdot \frac{n+1}{n} = \frac{1}{n \cdot n!}$$

Why this bound is the right one

Multiply the inequality by $n!$:

$$0 < n!\,R_n < \frac{1}{n}$$

The quantity $n!\,R_n$ is therefore squeezed strictly between $0$ and $\dfrac{1}{n}$, that is, strictly between $0$ and $1$ as soon as $n \geq 2$. It will be enough to prove that this number is an integer to reach the contradiction: that is exactly the program of the next chapter.

Classic pitfall

Confusing $\dfrac{1}{n \cdot n!}$ with $\dfrac{1}{(n+1)!}$, or settling for the crude bound $R_n < \dfrac{1}{(n+1)!} \cdot (n+1) = \dfrac{1}{n!}$. The latter gives $0 < n!\,R_n < 1$, which is in fact also enough for the contradiction; the version with $\dfrac{1}{n \cdot n!}$ is simply sharper.