How do you use the integral test to determine if Sigma1/sqrt(n+1) from [1,oo) is convergent or divergent?

Nov 14, 2017

The series is divergent

Explanation:

Apply the integral test only if the function $f \left(x\right)$ is continuous, positive and decreasing.

The function $f \left(x\right)$ is continuous as $\forall x \in \left[1 , \infty\right)$ as $\sqrt{1 + x}$ is continuous.

The function is positive as $\sqrt{1 + x}$ is positive in the interval $\left[1 , \infty\right]$

$f ' \left(x\right) = \left(\frac{1}{\sqrt{1 + x}}\right) ' = - \frac{1}{2} {\left(1 + x\right)}^{- \frac{3}{2}}$

Therefore,

$f ' \left(x\right) < 0 , \forall x \in \left[1 , \infty\right)$

So, we can apply the integral test.

Calculate the improper integral

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

First compute the indefinite integral

$\int \frac{\mathrm{dx}}{\sqrt{1 + x}} = \int {\left(1 + x\right)}^{- \frac{1}{2}} \mathrm{dx}$

$= {\left(1 + x\right)}^{- \frac{1}{2} + 1} / \left(- \frac{1}{2} + 1\right)$

$= 2 {\left(1 + x\right)}^{\frac{1}{2}}$

Therefore,

${\lim}_{t \to \infty} {\int}_{1}^{t} \frac{\mathrm{dx}}{\sqrt{1 + x}} = {\lim}_{t \to \infty} {\left[2 {\left(1 + x\right)}^{\frac{1}{2}}\right]}_{1}^{t}$

$= {\lim}_{t \to \infty} \left(2 \sqrt{1 + t} - 2 \sqrt{2}\right)$

$= + \infty$

As the improper integral is divergent , so the ${\sum}_{n = 1}^{\infty} \frac{1}{\sqrt{1 + n}}$ is divergent