Test the following series for convergence
Σ from n=1to ♾️ for [√n⁴+9 - √n⁴-9]
limn→∞n4+9−n4−91/n2=limn→∞18n2(n4+9+n4−9)==limn→∞18(1+9n4+1−9n4)=182=9∑n=1∞1n2 converges⇒∑n=1∞(n4+9−n4−9) 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)}\,\,convergesn→∞lim1/n2n4+9−n4−9=n→∞lim(n4+9+n4−9)18n2==n→∞lim(1+n49+1−n49)18=218=9∑n=1∞n21converges⇒∑n=1∞(n4+9−n4−9)converges
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