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Use theorem 9. 11 to determine the convergence or divergence of the p-series. 1 1 6 32 1 6 243 1 6 1024 1 6 3125. .

Sagot :

The given series is diverges according to the P Test.

According to the statement

we have given that the a series and we have to find that the those series is a converges or diverges.

So, For this purpose, we know that the

P-series test to say whether or not the series converges.

So, The given series is

[tex]1 + 1/\sqrt{32} + 1/\sqrt[3]{243 } + 1/\sqrt[4]{1024} + 1/\sqrt[5]{3125}[/tex]

And then

The value of  P becomes P = 5/6 < 1  when we find it using the p series test.

Hence, the p-series diverges

Because according to the p series test, The series converges if, and only if, the power satisfies p>1.

And diverges if, and only if, the power satisfies p<1.

So, The given series is diverges according to the P Test.

Learn more about Convergence and Divergence here

https://brainly.com/question/4730059

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Question:

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.

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