# proof of Cauchy’s root test

If for all $n\geq N$

 $\sqrt[n]{a_{n}}

then

 $a_{n}

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 ProofOfCauchysRootTest 2013-03-22 13:23:43 2013-03-22 13:23:43 mathwizard (128) mathwizard (128) 5 mathwizard (128) Proof msc 40A05