Question

Improper Integrals: Determine whether the following improper integrals converge or diverge.

${\int }_1^{\infty }\frac{1}{x^2}dx.$

Short answer

The series is a convergent series.

Step-by-step solution

Step 1. Given Information.

$Givenanintegral:{\int }_1^{\infty }\frac{1}{x^2}dx.$

Step 2. Solving the integral.

$$\begin{gathered} {\int }_1^{\infty }\frac{1}{x^2}dx \\ ={\left[\frac{-1}{x}\right]}_1^{\infty } \\ =\left[\frac{-1}{\infty }-\left(\frac{-1}{1}\right)\right] \\ =\left[0+1\right]=1. \\ Therefore,theseriesisaconvergentseries. \end{gathered}$$