proof of Cauchy’s root test

If for all $n\geq N$

\sqrt[n]{a_{n}}<k<1

then

a_{n}<k^{n}<1.

Since $\sum_{i=N}^{\infty}k^{i}$ converges so does $\sum_{i=N}^{\infty}a_{n}$ by the comparison test. If $\sqrt[n]{a_{n}}>1$ then by comparison with $\sum_{i=N}^{\infty}1$ the series is divergent. Absolute convergence in case of nonpositive $a_{n}$ can be proven in exactly the same way using $\sqrt[n]{|a_{n}|}$ .

Title	proof of Cauchy’s root test
Canonical name	ProofOfCauchysRootTest
Date of creation	2013-03-22 13:23:43
Last modified on	2013-03-22 13:23:43
Owner	mathwizard (128)
Last modified by	mathwizard (128)
Numerical id	5
Author	mathwizard (128)
Entry type	Proof
Classification	msc 40A05