]> App01, Natural base

# Natural Base

## Convergence of the sequence

By definition the natural base is (1.2.4.2)

 $e=\underset{n\to \infty }{\mathrm{lim}}{\left(1+\frac{1}{n}\right)}^{n}=2.71828....$ (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  nth  term of the sequence by

 ${a}_{n}={\left(1+\frac{1}{n}\right)}^{n}$ (A01.2)

and by use of the binomial expansion (1.1.2.20),

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

we obtain:

 ${a}_{n}=\sum _{k=0}^{n}{c}_{nk}=\sum _{k=0}^{n}\left(\begin{array}{l}n\\ k\end{array}\right)\frac{1}{{n}^{k}}$ (A01.4)

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

 $\begin{array}{l}{c}_{n0}={c}_{n1}=1\text{ }\text{for}\text{\hspace{0.17em}}\text{any}\text{ }n>0\\ \begin{array}{l}{c}_{nk}=\frac{n\left(n-1\right)...\left(n-k+1\right)}{k!\text{ }{n}^{k}}=\\ =\frac{1}{k!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)...\left(1-\frac{k-1}{n}\right)\end{array}\right\}\text{for}\text{\hspace{0.28em}}k>1\end{array}\right\}$ (A01.5)

From (A01.5) it follows that

 ${c}_{\left(n+1\right)k}>{c}_{nk}\text{ }\text{for}\text{\hspace{0.28em}}k>1$ (A01.6)

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

As a lower bound of the sequence one can use

 ${c}_{n0}+{c}_{n1}=2$ (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  cnk :

 ${c}_{nk}\le \frac{1}{{2}^{k-1}}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)...\left(1-\frac{k-1}{n}\right)\text{ }\text{for}\text{\hspace{0.28em}}k>1$ (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  cnk . Finally we obtain:

 $\begin{array}{l}{c}_{nk}<\frac{1}{{2}^{k-1}}\text{ }\text{for}\text{\hspace{0.28em}}\text{any}\text{\hspace{0.28em}}k>1,\text{\hspace{0.28em}}\text{yielding:}\\ {a}_{n}<2+\sum _{k=2}^{n}\frac{1}{{2}^{k-1}}=\\ =2+\frac{1}{2}+\frac{1}{{2}^{2}}+....+\frac{1}{{2}^{n-1}}<3\end{array}\right\}$ (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 (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)

 $e=\underset{n\to \infty }{\mathrm{lim}}{a}_{n}=\underset{n\to \infty }{\mathrm{lim}}\sum _{k=0}^{n}{c}_{nk}$ (A01.13)

exists, one could expect

 $e=\underset{n\to \infty }{\mathrm{lim}}{a}_{n}=\sum _{k=0}^{\infty }\underset{n\to \infty }{\mathrm{lim}}{c}_{nk}$ (A01.14)

Indeed, from (A01.5) it follows that

 $\underset{n\to \infty }{\mathrm{lim}}{c}_{nk}=\frac{1}{k!}\text{ }\text{for}\text{\hspace{0.28em}}k>0$ (A01.15)

in full agreement with (A01.12).