Two Roads to e | Real Analysis

Compare how fast compound growth

(1+\frac{1}{n})^n

and the series

\sum_{k=0}^{\infty} \frac{1}{k!}

approach

e

. The factorial series wins by a landslide.

Examples & Exercises

Prove that $e$ is irrational.

0 < q!\left(e - \sum_{k=0}^{q}\frac{1}{k!}\right) < 1

n = 10

: the series is already within

3\times 10^{-8}

e

Show proof

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.

Prove the remainder bound $R_n = \sum_{k>n}\frac{1}{k!} < \frac{1}{n\cdot n!}$ .

R_n < \frac{1}{(n+1)!}\left(1+\frac{1}{n+1}+\frac{1}{(n+1)^2}+\cdots\right)=\frac{1}{n\cdot n!}

n = 10

: actual remainder

\approx 2.73\times 10^{-8}

, bound

\approx 2.76\times 10^{-8}

Show proof

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

Show $\left(1+\frac{1}{n}\right)^n$ increases toward $e$ from below. (Monotonicity is sketched at idea level.)

\left(1+\frac{1}{n}\right)^n = \sum_{k=0}^{n}\frac{1}{k!}\prod_{j=0}^{k-1}\left(1-\frac{j}{n}\right) < e

n = 30

: the compounding road is still far below

e

Show proof

By the binomial theorem, $\left(1+\frac{1}{n}\right)^n = \sum_{k=0}^{n}\binom{n}{k}\frac{1}{n^k} = \sum_{k=0}^{n}\frac{1}{k!}(1)\left(1-\frac{1}{n}\right)\cdots\left(1-\frac{k-1}{n}\right)$ .
Each factor $\left(1-\frac{j}{n}\right) < 1$ , so $\left(1+\frac{1}{n}\right)^n < \sum_{k=0}^{n}\frac{1}{k!} < e$ .
As $n$ grows, the factors $\left(1-\frac{j}{n}\right)$ approach $1$ , so the sequence increases (the rigorous monotonicity proof is omitted — idea level).