Editing Root test (section)

==Examples==

''Example 1:'' 
:<math> \sum_{i=1}^\infty \frac{2^i}{i^9} </math>
Applying the root test and using the fact that <math> \lim_{n \to \infty} n^{1/n}=1,</math>
::<math> C = \lim_{n \to \infty}\sqrt[n]{\left|\frac{2^n}{n^9}\right|}= \lim_{n \to \infty}\frac{ \sqrt[n]{2^n} } { \sqrt[n]{n^9} } = \lim_{n \to \infty}\frac{ 2 }   {(n^{1/n})^9 }  = 2 </math> 
Since <math> C=2>1,</math>  the series diverges.<ref>{{cite book
 | first1= William |last1= Briggs|first2= Lyle|last2 = Cochrane
 | title= Calculus: Early Transcendentals
 | url= https://archive.org/details/calculusearlytra0000brig/page/n5/mode/2up
 | url-access= registration
 | publisher= Addison Wesley
 | year=2011
 }}  p. 571. </ref>

''Example 2:'' 
:<math>\sum_{n=0}^\infty \frac{1}{2^{\lfloor n/2 \rfloor}}= 1 + 1 + \frac12 + \frac12 + \frac14 + \frac14 + \frac18 + \frac18  + \ldots </math>
The root test shows convergence because
:: <math>r= \limsup_{n\to\infty}\sqrt[n]{|a_n|} = \limsup_{n\to\infty}\sqrt[2n]{|a_{2n}|} = \limsup_{n\to\infty}\sqrt[2n]{|1/2^n|}=\frac1\sqrt{2}<1.</math>
This example shows how the root test is stronger than the [[ratio test]]. The ratio test is inconclusive for this series as if <math>n</math> is even, <math>a_{n+1}/a_n = 1</math> while if <math>n</math> is odd, <math>a_{n+1}/a_n = 1/2</math>, therefore the limit <math>\lim_{n\to\infty} |a_{n+1}/a_n|</math> does not exist.