# How do you use the integral test to determine the convergence or divergence of Sigma 1/sqrtn from [1,oo)?

Jul 18, 2018

The series ${\sum}_{n = 1}^{\infty} \frac{1}{\sqrt{n}}$ is divergent.

#### Explanation:

Let $f \left(x\right) = \frac{1}{\sqrt{x}}$.

We have that:

1) $f \left(x\right) > 0$

2)${\lim}_{x \to \infty} f \left(x\right) = 0$

3) $\frac{\mathrm{df}}{\mathrm{dx}} = - \frac{1}{2 x \sqrt{x}} < 0$ so that $f \left(x\right)$ is strictly decreasing in $\left[1 , + \infty\right)$

4) $f \left(n\right) = \frac{1}{\sqrt{n}}$

Under these conditions the convergence of the series:

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

is equivalent to the convergence of the improper integral:

${\int}_{1}^{\infty} \frac{\mathrm{dx}}{\sqrt{x}}$

Evaluate the indefinite integral:

$\int \frac{\mathrm{dx}}{\sqrt{x}} = 2 \sqrt{x}$

and we can see that the integral:

${\int}_{1}^{\infty} \frac{\mathrm{dx}}{\sqrt{x}} = - 2 + 2 {\lim}_{x \to \infty} \sqrt{x}$

is not convergent.

Then also the series ${\sum}_{n = 1}^{\infty} \frac{1}{\sqrt{n}}$ is not convergent, and as $\frac{1}{\sqrt{n}} > 0$ we can conclude that it is divergent.