The Unapologetic Mathematician

Mathematics for the interested outsider

The Ratio and Root Tests

Now I want to bring out with two tests that will tell us about absolute convergence or (unconditional) divergence of an infinite series \sum_{k=0}^\infty a_k. As such they’ll tell us nothing about conditionally convergent series.

First is the ratio test. We take the ratio of one term in the series to the previous one and define the limits superior and inferior

\displaystyle R=\limsup\limits_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|
\displaystyle r=\liminf\limits_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|

Now if R<1 then the series converges absolutely. If r>1 then the series diverges. But if r\leq1\leq R the test fails and we get no result.

In the first case, pick x to be a number so that R<x<1. Then there is some N so that x is an upper bound for the sequence of ratios past N. For large enough n, this means


and so


Now if we set c=\frac{\left|a_N\right|}{x^N}, this tells us that |a_n|\leq cx^n. Then the comparison test with the geometric series tells us that \sum_{k=0}^\infty\left|a_k\right| converges.

On the other hand, if r>1 then eventually \left|a_{n+1}\right|>\left|a_n\right|, so the terms of the series are getting bigger and bigger and bigger. But this would throw a monkey wrench into Cauchy’s condition for convergence of the series.

As for the root test, we will consider the sequence \sqrt[n]{\left|a_n\right|} and define


If \rho<1 then the series converges absolutely. If \rho>1 then the series diverges. And if \rho=1 the test is inconclusive.

In the first case, as we did for the ratio test, pick x so that \rho<x<1. Then above some N we have \left|a_n\right|<x^n and the comparison test works straight away. On the other hand, if \rho>1 then \left|a_n\right|>1 infinitely often, and Cauchy’s criterion falls apart again.

May 5, 2008 Posted by | Analysis, Calculus | 5 Comments