Answer to Question #206165 in Real Analysis for Rajkumar

Question

Test the following series for convergence

∑

∞

n=1 [✓(n^4 +9) -✓(n^4 -9)]

Answer 1

$$\displaystyle\sum_{n=1}^{\infin}(\sqrt{n^4+9}-\sqrt{n^4-9})$$

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

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

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

Use the Limit Comparison Test with

$$a_n=\dfrac{1}{\sqrt{n^4+9}+\sqrt{n^4-9}}, b_n=\dfrac{1}{n^2}$$

and obtain

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

$$=\lim\limits_{n\to\infin}\dfrac{1}{\sqrt{1+\dfrac{9}{n^4}}+\sqrt{1-\dfrac{9}{n^4}}}=\dfrac{1}{2}>0$$

Since

$$\displaystyle\sum_{n=1}^{\infin}b_n=\displaystyle\sum_{n=1}^{\infin}\dfrac{1}{n^2}$$

is convergent ($p$ -series with $p=2$ ), the series $\displaystyle\sum_{n=1}^{\infin}\dfrac{1}{\sqrt{n^4+9}+\sqrt{n^4-9}}$ converges by the Limit Comparison Test.

Therefore the given series $\displaystyle\sum_{n=1}^{\infin}(\sqrt{n^4+9}-\sqrt{n^4-9})$ is convergent by the Limit Comparison Test.