/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 32 Determine the convergence or div... [FREE SOLUTION] | 91Ó°ÊÓ

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Determine the convergence or divergence of the following series. $$\sum_{k=1}^{\infty} 2 k^{-3 / 2}$$

Short Answer

Expert verified
Answer: The series converges.

Step by step solution

01

Identify the p-value

First, rewrite the given series in the context of the p-series test: $$\sum_{k=1}^{\infty} 2 k^{-3 / 2} = 2 \sum_{k=1}^{\infty} k^{-3 / 2} = 2 \sum_{k=1}^{\infty} \frac{1}{k^{3 / 2}}$$ We can see that our p-value is 3/2.
02

Apply the P-test

Now that we have identified the p-value as 3/2, we can apply the p-test. Since 3/2 > 1, the p-test indicates that the series converges.
03

Conclusion

Using the p-test, the given series $$\sum_{k=1}^{\infty} 2 k^{-3 / 2}$$ converges.

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