Answer to Question #211989 in Calculus for Vikas Kumar

Question

[√n^4+9-√n^4-9]∞∑n=0 test the congervence

Answer 1

$$\displaystyle\sum_{n=2}^{\infin}(\sqrt{n^4+9}-\sqrt{n^4-9})$$

$$=\displaystyle\sum_{n=2}^{\infin}\dfrac{(\sqrt{n^4+9}-\sqrt{n^4-9})(\sqrt{n^4+9}+\sqrt{n^4-9})}{\sqrt{n^4+9}+\sqrt{n^4-9}}$$

$$=\displaystyle\sum_{n=2}^{\infin}\dfrac{n^4+9-n^4+9}{\sqrt{n^4+9}+\sqrt{n^4-9}}$$

$$=18\displaystyle\sum_{n=2}^{\infin}\dfrac{1}{\sqrt{n^4+9}+\sqrt{n^4-9}}$$

The p-series $\displaystyle\sum_{n=2}^{\infin}\dfrac{1}{n^2}$ converges since $p=2>1.$

$$\lim\limits_{n\to \infin}\dfrac{\dfrac{1}{\sqrt{n^4+9}+\sqrt{n^4-9}}}{\dfrac{1}{n^2}}$$

$$=\lim\limits_{n\to \infin}\dfrac{n^2}{\sqrt{n^4+9}+\sqrt{n^4-9}}$$

$$=\lim\limits_{n\to \infin}\dfrac{\dfrac{n^2}{n^2}}{\dfrac{\sqrt{n^4+9}+\sqrt{n^4-9}}{n^2}}$$

$$=\dfrac{1}{1+1}=\dfrac{1}{2}<\infin$$

Therefore the given series $\displaystyle\sum_{n=2}^{\infin}(\sqrt{n^4+9}-\sqrt{n^4-9})$ converges by the Limit Comparison Test.