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[parent] application of Cauchy criterion for convergence (Example)

Without using the methods of the entry determining series convergence, we show that the real-term series $$\sum_{n=0}^\infty\frac{1}{n!} \;=\; 1+\frac{1}{1!}+\frac{1}{2!}+\ldots$$ is convergent by using Cauchy criterion for convergence, being in force in $\mathbb{R}$ equipped with the usual absolute value $|.|$ as norm.

Let $\varepsilon$ be an arbitrary positive number. For any positive integer $n$ , we have $$\frac{1}{n!} \;\leqq\; \frac{1}{1\cdot2\cdot2\cdots2} \;=\; \frac{1}{2^{n-1}},$$ whence we can estimate as follows.

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

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.




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See Also: real number, geometric series, logarithmus binaris


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Cross-references: converges, implies, inequality, integer, number, positive, absolute value, Cauchy criterion for convergence, convergent, series, determining series convergence

This is version 5 of application of Cauchy criterion for convergence, born on 2009-10-05, modified 2009-11-17.
Object id is 11936, canonical name is ApplicationOfCauchyCriterionForConvergence.
Accessed 246 times total.

Classification:
AMS MSC40A05 (Sequences, series, summability :: Convergence and divergence of infinite limiting processes :: Convergence and divergence of series and sequences)

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