proof of Cauchy’s root test
If for all n≥N
n√an<k<1 |
then
an<kn<1. |
Since ∑∞i=Nki converges so does ∑∞i=Nan by the comparison test. If n√an>1 then by comparison with ∑∞i=N1 the series is divergent. Absolute convergence in case of nonpositive an can be proven in exactly the same way using n√|an|.
Title | proof of Cauchy’s root test![]() |
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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 |