Question #314405

Test the following series for convergence


Σ from n=1to ♾️ for [√n⁴+9 - √n⁴-9]

1
Expert's answer
2022-03-23T14:22:42-0400

limnn4+9n491/n2=limn18n2(n4+9+n49)==limn18(1+9n4+19n4)=182=9n=11n2convergesn=1(n4+9n49)converges\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


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