Respuesta :
Answer:
Divergent.
Step-by-step explanation:
We have been given an integral [tex]\int\limits^\infty _1 {\frac{1}{x^{0.999}} \, dx[/tex]. We are asked to determine whether the given integral diverges or converges.
[tex]\int _1^{\infty }\:\:\:\frac{1}{x^{0.999}}\:\:\:dx[/tex]
[tex]\int _1^{\infty }\:\:\:\frac{1}{x^{0.999}}\:\:\:dx=\int _1^{\infty }\:\:\:x^{-0.999}\:\:\:dx[/tex]
[tex]\int _1^{\infty }\:\:\:x^{-0.999}\:\:\:dx=\frac{x^{-0.999+1}}{-0.999+1}[/tex]
[tex]\int _1^{\infty }\:\:\:x^{-0.999}\:\:\:dx=\frac{x^{0.001}}{0.001}[/tex]
[tex]\int _1^{\infty }\:\:\:x^{-0.999}\:\:\:dx=1000x^{0.001}[/tex]
Let us compute the boundaries.
[tex]1000(\infty)^{0.001}=\infty[/tex]
[tex]1000(1)^{0.001}=1000[/tex]
Since [tex]\infty-1000[/tex] is not a finite number, therefore, the given integral diverges.
