Tuesday, 6 November 2018

sequences and series - Convergence of suminftyn=1frac2sqrtn+2



In the Comparison Tests section of my textbook, I am tasked with determining the convergence of the series n=12n+2



I will argue the series diverges using the "Limit" Comparison test.




Consider an=2n+2 and bn=1n Note: The latter series bn is a divergent p-series with p=12.



If lim where c>0, then either both a_n and b_n converge, or both a_n and b_n diverge.



Thus \lim_{n\rightarrow\infty}\frac{a_n}{b_n}=\lim_{n\rightarrow\infty}\frac{2\sqrt{n}}{\sqrt{n}+2}=\lim_{n\rightarrow\infty}\frac{2}{1+\frac{2}{\sqrt{n}}}=2



Since the above limit existed as a finite value c=2>0 and since \sum b_n diverges (p-series with p=\frac{1}{2}):





\sum^{\infty}_{n=1}\frac{2}{\sqrt{n}+2}\space\text{diverges}




However, from the beginning I can see the asymptotic equivalence: \frac{2}{\sqrt{n}+2}\sim\frac{1}{\sqrt{n}} Why do I need the limit comparison test when I can deduce the result quickly upon inspection? Thanks in advance!


Answer



Note that \frac{2}{\sqrt{n}+2}\geq \frac{1}{\sqrt{n}}


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