Answer to Question #314405 in Real Analysis for Pankaj

Question

Test the following series for convergence

Σ from n=1to ♾️ for [√n⁴+9 - √n⁴-9]

Answer 1

$\underset{n\rightarrow \infty}{\lim}\frac{\sqrt{n^4+9}-\sqrt{n^4-9}}{1/n^2}=\underset{n\rightarrow \infty}{\lim}\frac{18n^2}{\left( \sqrt{n^4+9}+\sqrt{n^4-9} \right)}=\\=\underset{n\rightarrow \infty}{\lim}\frac{18}{\left( \sqrt{1+\frac{9}{n^4}}+\sqrt{1-\frac{9}{n^4}} \right)}=\frac{18}{2}=9\\\sum_{n=1}^{\infty}{\frac{1}{n^2}}\,\,converges\Rightarrow \sum_{n=1}^{\infty}{\left( \sqrt{n^4+9}-\sqrt{n^4-9} \right)}\,\,converges$