# How do you test the series Sigma (sqrt(n+2)-sqrtn)/n from n is [1,oo) for convergence?

Feb 5, 2017

The series:

${\sum}_{n = 1}^{\infty} \frac{\sqrt{n + 2} - \sqrt{n}}{n}$

is convergent.

#### Explanation:

Rationalize the numerator of ${a}_{n}$:

a_n = (sqrt(n+2)-sqrt(n))/n = ((sqrt(n+2)-sqrt(n))/n)((sqrt(n+2)+sqrt(n))/(sqrt(n+2)+sqrt(n))) = (n+2-n)/(n(sqrt(n+2)+sqrt(n))) = 2/(n(sqrt(n+2)+sqrt(n))

Now if we decrease the denominator we obtain a sequence that is greater:

${a}_{n} = \frac{2}{n \left(\sqrt{n + 2} + \sqrt{n}\right)} < \frac{2}{n \left(\sqrt{n} + \sqrt{n}\right)} = \frac{2}{2 n \sqrt{n}} = \frac{1}{n} ^ \left(\frac{3}{2}\right)$

We know that the series:

${\sum}_{n = 1}^{\infty} \frac{1}{n} ^ \left(\frac{3}{2}\right)$

is convergent based on the p-series test, so the series:

${\sum}_{n = 1}^{\infty} \frac{\sqrt{n + 2} - \sqrt{n}}{n}$

is also convergent.