test ∑n=1 ∞ [✓n^4 +9 - ✓n^4 -9] is convergent or not
Σn=1∞(n4+9−n4−9)==Σn=1∞(n4+9−n4−9)(n4+9+n4−9)(n4+9+n4−9)==Σn=1∞18n4+9+n4−918n4+9+n4−9↔182n2=9n2\Sigma_{n=1}^{\infty}(\sqrt{n^4+9}-\sqrt{n^4-9})=\\ =\Sigma_{n=1}^{\infty}\frac{(\sqrt{n^4+9}-\sqrt{n^4-9})(\sqrt{n^4+9}+\sqrt{n^4-9})}{(\sqrt{n^4+9}+\sqrt{n^4-9})}=\\ =\Sigma_{n=1}^{\infty}\frac{18}{\sqrt{n^4+9}+\sqrt{n^4-9}}\\ \frac{18}{\sqrt{n^4+9}+\sqrt{n^4-9}}\leftrightarrow\frac{18}{2n^2}=\frac{9}{n^2}Σn=1∞(n4+9−n4−9)==Σn=1∞(n4+9+n4−9)(n4+9−n4−9)(n4+9+n4−9)==Σn=1∞n4+9+n4−918n4+9+n4−918↔2n218=n29
Σn=1∞9n2\Sigma_{n=1}^{\infty}\frac{9}{n^2}Σn=1∞n29 is convergent
Σn=1∞(n4+9−n4−9)\Sigma_{n=1}^{\infty}(\sqrt{n^4+9}-\sqrt{n^4-9})Σn=1∞(n4+9−n4−9) is convergent
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