# application of Cauchy criterion for convergence

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

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

whence we can as follows.

 $\displaystyle\left|\frac{1}{(n\!+\!1)!}+\ldots+\frac{1}{(n\!+\!p)!}\right|$ $\displaystyle\;=\;\frac{1}{(n\!+\!1)!}+\ldots+\frac{1}{(n\!+\!p)!}$ $\displaystyle\;\leqq\;\frac{1}{2^{n}}+\ldots+\frac{1}{2^{n+p-1}}$ $\displaystyle\;=\;\frac{1}{2^{n}}\left(1+\frac{1}{2}+\ldots+\frac{1}{2^{p-1}}\right)$ $\displaystyle\;=\;\frac{1}{2^{n}}\cdot\frac{1-(1/2)^{p}}{1-1/2}$ $\displaystyle\;<\;\frac{1}{2^{n-1}}\;<\;\varepsilon$

The last inequality  is true for all positive integers $p$, when  $n\;>\;1-\mbox{lb}\,{\varepsilon}$.  Thus the Cauchy criterion implies that the series converges.

Title application of Cauchy criterion for convergence ApplicationOfCauchyCriterionForConvergence 2013-03-22 19:03:22 2013-03-22 19:03:22 pahio (2872) pahio (2872) 9 pahio (2872) Example msc 40A05 RealNumber GeometricSeries LogarithmusBinaris NapiersConstant