The number e

We are not always able to express the limit explicitly (i.e. the real number itself), in fact some real numbers are defined as limits of sequences. For instance consider

$$a_n={(1+\frac{1}{n})}^n$$

We shall show now that {a_n} is increasing and bounded above. Therefore it is convergent. The real number which is the limit of this sequence is denoted by e and called the Euler number.

Proposition 3.4.1. The sequence {a_n} = (1 + 1/n)ⁿ is strictly increasing.

$$\begin{array}{rl}{a}_n& ={(1+\frac{1}{n})}^n=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})\frac{1}{n^k}=\sum _{k=0}^n\frac{1}{k!}\frac{n(n-1)\dots (n-k+1)}{n^k}\\ & =\sum _{k=0}^n\frac{1}{k!}\frac{n}{n}\frac{n-1}{n}\dots \frac{n-k+1}{n}\\ & =\sum _{k=0}^n\frac{1}{k!}\cdot 1\cdot (1-\frac{1}{n})\dots (1-\frac{k-1}{n})\end{array}\text{\hspace{1em} 3.3.1}$$

similarly

$$a_{n+1}=\sum _{k=0}^{n+1}\frac{1}{k!}\cdot 1\cdot (1-\frac{1}{n+1})\dots (1-\frac{k-1}{n+1})\text{\hspace{1em} 3.3.2}$$

We know that

$$(1-\frac{1}{n})<(1-\frac{1}{n+1}),\dots ,(1-\frac{k-1}{n})<(1-\frac{k-1}{n+1})$$

so each summand of 3.3.1 is smaller than the corresponding term in (3.3.2). The latter sum, moreover, contains an additional positive summand labelled by k = n + 1. Therefore a_n < a_n+1 for each n.  □

Proposition 3.4.2. The sequence a_n is bounded; precisely

Proof. Since a₁ = 2, and the sequence is strictly monotone by the previous property, we have a_n > 2, ∀n > 1. Let us show that a_n > 3, ∀n > 1. By (3.3.1) and observing that

we have that

$$\begin{array}{rl}{a}_n& =\sum _{k=0}^n\frac{1}{k!}\cdot 1\cdot (1-\frac{1}{n})\cdots (1-\frac{k-1}{n})<\sum _{k=0}^n\frac{1}{k!}\\ & =1+\sum _{k=1}^n\frac{1}{k!}\le 1+\sum _{k=1}^n\frac{1}{2^{k-1}}=1+\sum _{k=0}^{n-1}\frac{1}{2^k}\end{array}$$

the majorative sum is that of a geometric progression that we know to be

$$\sum _{k=0}^{n-1}\frac{1}{2^k}=\frac{1-\frac{1}{2^n}}{1-\frac{1}{2}}=2(1-\frac{1}{2^n})<2$$

We conclude that a_n < 3.  □

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