application of Cauchy criterion for convergence

Without using the methods of the entry determining series convergence, we show that the real-term series

$$\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}=\mathrm{\hspace{0.33em}1}+\frac{1}{1!}+\frac{1}{2!}+\mathrm{\dots }$$

is convergent by using Cauchy criterion for convergence, being in in $ℝ$ equipped with the usual absolute value $|.|$ as http://planetmath.org/node/1604norm.

Let $\epsilon$ be an arbitrary positive number.  For any positive integer $n$, we have

$$\frac{1}{n!}\leqq \frac{1}{1\cdot 2\cdot 2\mathrm{\cdots }2}=\frac{1}{2^{n-1}},$$

whence we can as follows.

$\left|{\displaystyle \frac{1}{(n+1)!}}+\mathrm{\dots }+{\displaystyle \frac{1}{(n+p)!}}\right|$ $\mathrm{\hspace{0.33em}}={\displaystyle \frac{1}{(n+1)!}}+\mathrm{\dots }+{\displaystyle \frac{1}{(n+p)!}}$

$\mathrm{\hspace{0.33em}}\leqq {\displaystyle \frac{1}{2^n}}+\mathrm{\dots }+{\displaystyle \frac{1}{2^{n+p-1}}}$

$\mathrm{\hspace{0.33em}}={\displaystyle \frac{1}{2^n}}\left(1+{\displaystyle \frac{1}{2}}+\mathrm{\dots }+{\displaystyle \frac{1}{2^{p-1}}}\right)$

$\mathrm{\hspace{0.33em}}={\displaystyle \frac{1}{2^n}}\cdot {\displaystyle \frac{1-{(1/2)}^p}{1-1/2}}$

The last inequality is true for all positive integers $p$, when  $n>\mathrm{\hspace{0.33em}1}-\text{lb}\epsilon$.  Thus the Cauchy criterion implies that the series converges.

Title	application of Cauchy criterion for convergence
Canonical name	ApplicationOfCauchyCriterionForConvergence
Date of creation	2013-03-22 19:03:22
Last modified on	2013-03-22 19:03:22
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	9
Author	pahio (2872)
Entry type	Example
Classification	msc 40A05
Related topic	RealNumber
Related topic	GeometricSeries
Related topic	LogarithmusBinaris
Related topic	NapiersConstant