Natural Base

Convergence of the sequence

By definition the natural base is (1.2.4.2)

$e = \lim_{n \to \infty} {(1 + \frac{1}{n})}^{n} = 2.71828....$	(A01.1)

According to a theorem, a sufficient and necessary condition for a monotonic (increasing or decreasing) sequence to converge, is the existence of two bounds (upper and lower), bounding all the terms.

Without proving the theorem, we'll use it, to show that the limit (A01.1) exists.

By denoting the n^th term of the sequence by

$a_{n} = {(1 + \frac{1}{n})}^{n}$	(A01.2)

and by use of the binomial expansion (1.1.2.20),

${\begin{cases} {(a + b)}^{n} = \sum_{k = 0}^{n} (\begin{array}{l} n \\ k \end{array}) a^{n - k} b^{k} \\ (\begin{array}{l} n \\ k \end{array}) = \frac{n!}{k! (n - k)!} \end{cases}}$	(A01.3)

we obtain:

$a_{n} = \sum_{k = 0}^{n} c_{n k} = \sum_{k = 0}^{n} (\begin{array}{l} n \\ k \end{array}) \frac{1}{n^{k}}$	(A01.4)

According to (A01.4) each term a_n of the sequence is represented by a sum of n+1 c_nk terms (with k=0,1,...,n). Simple calculation shows that

$\begin{array}{l} c_{n 0} = c_{n 1} = 1 for any n > 0 \\ \begin{array}{l} c_{n k} = \frac{n (n - 1) ... (n - k + 1)}{k! n^{k}} = \\ = \frac{1}{k!} (1 - \frac{1}{n}) (1 - \frac{2}{n}) ... (1 - \frac{k - 1}{n}) \end{array}} for k > 1 \end{array}}$	(A01.5)

From (A01.5) it follows that

$c_{(n + 1) k} > c_{n k} for k > 1$	(A01.6)

and also that for n+1 , an additional positive term $c_{(n + 1) (k = n + 1)}$ is added to the a_n . Therefore the sequence (A01.2) is increasing.

As a lower bound of the sequence one can use

$c_{n 0} + c_{n 1} = 2$	(A01.7)

For proving the convergence of the sequence (A01.2), we need to find an upper bound. For k>1 , the substitution of 2^(k-1) for k! in (A01.5), could only increase the values of c_nk :

$c_{n k} \leq \frac{1}{2^{k - 1}} (1 - \frac{1}{n}) (1 - \frac{2}{n}) ... (1 - \frac{k - 1}{n}) for k > 1$	(A01.8)

Furthermore, all the expressions of (A01.8) in brackets, that include n , are <1 and therefore their substitution by unity, increases the terms of c_nk . Finally we obtain:

$\begin{array}{l} c_{n k} < \frac{1}{2^{k - 1}} for any k > 1, yielding: \\ a_{n} < 2 + \sum_{k = 2}^{n} \frac{1}{2^{k - 1}} = \\ = 2 + \frac{1}{2} + \frac{1}{2^{2}} + .... + \frac{1}{2^{n - 1}} < 3 \end{array}}$	(A01.9)

The last inequality in (A01.9) was obtained from the geometric series (1.3.5.19):

$\sum_{n = 0}^{\infty} \frac{1}{2^{n}} = \frac{1}{1 - \frac{1}{2}} = 2$	(A01.10)

which shows that the sequence (A01.2) converges and

$2 < e < 3$	(A01.11)

Consistency with the series

By use of the Maclaurin series, we found a way (1.3.5.9) to calculate e with the series

$e = 1 + \sum_{n = 1}^{\infty} \frac{1}{n!}$	(A01.12)

This expression (A01.12) should be consistent with the definition of the natural base (A01.1). If the limit (A01.4)

$e = \lim_{n \to \infty} a_{n} = \lim_{n \to \infty} \sum_{k = 0}^{n} c_{n k}$	(A01.13)

exists, one could expect

$e = \lim_{n \to \infty} a_{n} = \sum_{k = 0}^{\infty} \lim_{n \to \infty} c_{n k}$	(A01.14)

Indeed, from (A01.5) it follows that

$\lim_{n \to \infty} c_{n k} = \frac{1}{k!} for k > 0$	(A01.15)

in full agreement with (A01.12).

Math Animated™ Mathematical Introduction for Physics and Engineering by Samuel Dagan (Copyright © 2007-2020)

Appendix 01

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Natural Base

Convergence of the sequence

Consistency with the series

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Math Animated™
Mathematical Introduction for Physics and Engineering
by Samuel Dagan (Copyright © 2007-2020)