Answer to Question #205070 in Real Analysis for Rajkumar

Question

Test the following series for convergence 

∑
n=1
∞ [✓n^4+9 -✓n^4-9]

Answer 1

$\sqrt{n^4 +9}-\sqrt{n^4 -9}\\ \text{Let} a _n=\sqrt{n^4 +9}-\sqrt{n^4 -9}\\ \text{Multiplying and dividing } a _n \text{by } \sqrt{n^4 +9}+\sqrt{n^4 -9}.\\ a_n=\frac{18}{\sqrt{n^4 +9}+\sqrt{n^4 -9}}\\ Then,\\ \frac{18}{\sqrt{n^4 +9}+\sqrt{n^4 -9}}= \frac{18}{n^2(\sqrt{1 +\frac{9}{n^4}}+\sqrt{1-\frac{9}{n^4}})}\leq\frac{18}{n^2} \\ \text{Let} \\ b_n=\frac{18}{n^2}.\\ \text{Since, for all n,} 0≤a_n≤b_n.\\ and ∑ _{i=1}^∞b_n =∑ _{i=1}^∞\frac{18}{n^2} (\text{Converges by p-test})\\ \text{By comparision test,}\\ \sum a_n \text{converges.}$