Two Roads to e

Compare how fast compound growth (1+1n)n(1+\frac{1}{n})^n and the series ∑k=0∞1k!\sum_{k=0}^{\infty} \frac{1}{k!} approach ee. The factorial series wins by a landslide.

e=∑k=0∞1k!e=\sum_{k=0}^{\infty}\frac{1}{k!}
e=lim⁡n→∞(1+1n)ne=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n

Examples & Exercises

Prove that ee is irrational.

0<q!(e−∑k=0q1k!)<10 < q!\left(e - \sum_{k=0}^{q}\frac{1}{k!}\right) < 1
n=10n = 10: the series is already within 3×10−83\times 10^{-8} of ee.
Show proof
  1. First, 2<e<32 < e < 3: e=∑k=0∞1k!>1+1=2e = \sum_{k=0}^{\infty}\frac{1}{k!} > 1+1 = 2, and since k!≥2k−1k! \ge 2^{k-1} for k≥1k \ge 1, e<1+1+12+14+⋯=3e < 1+1+\frac{1}{2}+\frac{1}{4}+\cdots = 3. So if e=pqe = \frac{p}{q} then q≥2q \ge 2.
  2. q!e=(q−1)! pq!e = (q-1)!\,p is an integer, and so is ∑k=0qq!k!\sum_{k=0}^{q}\frac{q!}{k!}. Put T=q!(e−∑k=0q1k!)T = q!\left(e - \sum_{k=0}^{q}\frac{1}{k!}\right); then T=1q+1+1(q+1)(q+2)+⋯T = \frac{1}{q+1}+\frac{1}{(q+1)(q+2)}+\cdots, so 0<T<1q+1(1+1q+1+⋯ )=1q≤12<10 < T < \frac{1}{q+1}\left(1+\frac{1}{q+1}+\cdots\right) = \frac{1}{q} \le \frac{1}{2} < 1.
  3. But q!e=(integer)+Tq!e = \text{(integer)} + T with 0<T<10 < T < 1 is not an integer — a contradiction. Hence ee is irrational.

Prove the remainder bound Rn=∑k>n1k!<1n⋅n!R_n = \sum_{k>n}\frac{1}{k!} < \frac{1}{n\cdot n!}.

Rn<1(n+1)!(1+1n+1+1(n+1)2+⋯ )=1n⋅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=10n = 10: actual remainder ≈2.73×10−8\approx 2.73\times 10^{-8}, bound ≈2.76×10−8\approx 2.76\times 10^{-8}.
Show proof
  1. Rn=1(n+1)!+1(n+2)!+⋯=1(n+1)!(1+1n+2+1(n+2)(n+3)+⋯ )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).
  2. The bracket is less than 1+1n+1+1(n+1)2+⋯=n+1n1+\frac{1}{n+1}+\frac{1}{(n+1)^2}+\cdots = \frac{n+1}{n} (geometric series).
  3. Hence Rn<1(n+1)!⋅n+1n=1n⋅n!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 (1+1n)n\left(1+\frac{1}{n}\right)^n increases toward ee from below.

(1+1n)n=∑k=0n1k!∏j=0k−1(1−jn)<e\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
sn+1−sn≥(an+1,2−an,2)+an+1,n+1>0s_{n+1} - s_n \ge (a_{n+1,2} - a_{n,2}) + a_{n+1,n+1} > 0
n=30n = 30: the compounding road is still far below ee.
Show proof
  1. Put sn=(1+1n)ns_n = \left(1+\frac{1}{n}\right)^n. By the binomial theorem sn=∑k=0n(nk)1nks_n = \sum_{k=0}^{n}\binom{n}{k}\frac{1}{n^k}, and since (nk)1nk=1k!∏j=0k−1(1−jn)\binom{n}{k}\frac{1}{n^k} = \frac{1}{k!}\prod_{j=0}^{k-1}\left(1-\frac{j}{n}\right) (empty product =1= 1 for k=0k = 0), sn=∑k=0nan,ks_n = \sum_{k=0}^{n} a_{n,k} with an,k=1k!∏j=0k−1(1−jn)a_{n,k} = \frac{1}{k!}\prod_{j=0}^{k-1}\left(1-\frac{j}{n}\right).
  2. Termwise comparison: for fixed k≥1k \ge 1 and 0<j≤k−10 < j \le k-1, (1−jn)<(1−jn+1)\left(1-\frac{j}{n}\right) < \left(1-\frac{j}{n+1}\right) (larger nn subtracts less), so an,k≤an+1,ka_{n,k} \le a_{n+1,k} — and for k≥2k \ge 2 the factor j=1j = 1 is strictly smaller, giving an,k<an+1,ka_{n,k} < a_{n+1,k}. All terms are positive.
  3. Hence sn+1=∑k=0n+1an+1,k=∑k=0nan+1,k+an+1,n+1≥∑k=0nan,k+(an+1,2−an,2)⏟> 0+an+1,n+1⏟> 0>sns_{n+1} = \sum_{k=0}^{n+1} a_{n+1,k} = \sum_{k=0}^{n} a_{n+1,k} + a_{n+1,n+1} \ge \sum_{k=0}^{n} a_{n,k} + \underbrace{(a_{n+1,2} - a_{n,2})}_{>\,0} + \underbrace{a_{n+1,n+1}}_{>\,0} > s_n — termwise increase plus one extra term. So sns_n is strictly increasing.
  4. Each factor (1−jn)<1\left(1-\frac{j}{n}\right) < 1, so an,k<1k!a_{n,k} < \frac{1}{k!} and sn<∑k=0n1k!<∑k=0∞1k!=es_n < \sum_{k=0}^{n}\frac{1}{k!} < \sum_{k=0}^{\infty}\frac{1}{k!} = e — sns_n increases from below toward ee. Every step is complete.

8. Power Series