# How do you use the Integral test on the infinite series sum_(n=1)^oo1/sqrt(n+4) ?

Sep 20, 2014

Since the integral

${\int}_{1}^{\infty} \frac{1}{\sqrt{x + 4}} \mathrm{dx}$

diverges, the series

${\sum}_{n = 1}^{\infty} \frac{1}{\sqrt{n + 4}}$

also diverges by Integral Test.

Let us evaluate the integral.

${\int}_{1}^{\infty} \frac{1}{\sqrt{x + 4}} \mathrm{dx}$

by the definition of improper integral,

$= {\lim}_{t \to \infty} {\int}_{1}^{t} \frac{1}{\sqrt{x + 4}} \mathrm{dx}$

by taking the antiderivative,

$= 2 {\lim}_{t \to \infty} {\left[\sqrt{x + 4}\right]}_{1}^{t}$

$= 2 {\lim}_{t \to \infty} \left(\sqrt{t + 4} - \sqrt{5}\right) = \infty$