Question

Find whether the following series are convergent or not

ii. ∞Σn=1 (√(n^2+3) - √(n^2-3)/ √n

Answer 1

$$\displaystyle\sum_{n=2}^{\infin}\dfrac{\sqrt{n^2+3}-\sqrt{n^2-3}}{n}$$

$$=\displaystyle\sum_{n=2}^{\infin}\dfrac{n^2+3-n^2+3}{n(\sqrt{n^2+3}+\sqrt{n^2-3})}$$

$$=\displaystyle\sum_{n=2}^{\infin}\dfrac{6}{n(\sqrt{n^2+3}+\sqrt{n^2-3})}$$

Use Limit Comparison Test

$$\lim\limits_{n\to\infin}\dfrac{a_n}{b_n}=\lim\limits_{n\to\infin}\dfrac{\dfrac{6}{n(\sqrt{n^2+3}+\sqrt{n^2-3})}}{\dfrac{1}{n^2}}=6,$$

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

Therefore the series $\displaystyle\sum_{n=2}^{\infin}\dfrac{\sqrt{n^2+3}-\sqrt{n^2-3}}{n}$ is convergent by Limit Comparison Test.