Use Theorem 9.11 to determine the convergence or divergence of the p-series. 1 + 1/32 squareroot + 1/243 squareroot 3 + 1/1024 squareroot 4 + 1/3125 squareroot 5 + p = ______
converges
diverges

Respuesta :

Answer:

Diverges

Step-by-step explanation:

since the value of  P = 5/6 < 1  the p-series diverges

summation of 1/n^5/6  at n = 1 and infinity

attached below is the detailed solution using Theorem 9.11 to determine the convergence or divergence of the p-series

Ver imagen batolisis

The value of the p in the p-series is less than one. Then the p-series is a divergence.

What is p-series?

It is an infinite type of series.

The p-series is given below.

[tex]1 + \dfrac{1}{\sqrt[6]{32} } +\dfrac{1}{\sqrt[6]{243} }+\dfrac{1}{\sqrt[6]{1024} }+\dfrac{1}{\sqrt[6]{3125} }[/tex]

The series can be written as

[tex]\dfrac{1}{1^{5/6}}+\dfrac{1}{2^{5/6}}+\dfrac{1}{3^{5/6}}+\dfrac{1}{4^{5/6}}+\dfrac{1}{5^{5/6}}[/tex]

Then the sum will be

[tex]\rm \Rightarrow \Sigma _{n=0} ^{\infty} \dfrac{1}{n^{5/6}} \\\\\\\Rightarrow P = \dfrac{5}{6} < 1[/tex]

Since the value of the p is <1. Then the series is divergence.

More about the p-series link is given below.

https://brainly.com/question/24249275