Chapter 7: Q 19. (page 631)

In Example 1 we used the comparison test to show that the series $\sum_{k = 1}^{\infty} \frac{k^{\frac{3}{2}} - k - 1}{5 k^{3} + 3}$ converges. Use the limit comparison test to prove the same result.

Short Answer

Expert verified

$The value of \lim_{k \to \infty} \frac{a_{k}}{b_{k}} = \frac{1}{5}; which is non zero finite number . The series \sum_{k = 1}^{\infty} b_{k} = \sum_{k = 1}^{\infty} \frac{1}{k^{\frac{3}{2}}} is convergent by p - series test . Therefore, the series \sum_{k = 1}^{\infty} a_{k} is also convergent . Hence, the series \sum_{k = 1}^{\infty} \frac{k^{\frac{3}{2}} - k - 1}{5 k^{3} + 3} is convergent .$

Step by step solution

Step 1. Given information is:

$\sum_{k = 1}^{\infty} \frac{k^{\frac{3}{2}} - k - 1}{5 k^{3} + 3}$

Step 2. Finding the term ∑k=1∞ bk

$The terms of the series \sum_{k = 1}^{\infty} \frac{k^{\frac{3}{2}} - k - 1}{5 k^{3} + 3} are positive . The series \sum_{k = 1}^{\infty} b_{k} for the series \sum_{k = 1}^{\infty} \frac{k^{\frac{3}{2}} - k - 1}{5 k^{3} + 3} is given by : \sum_{k = 1}^{\infty} b_{k} = \sum_{k = 1}^{\infty} \frac{k^{\frac{3}{2}}}{k^{3}} (Dominant term of numerator and denominator) = \sum_{k = 1}^{\infty} \frac{1}{k^{\frac{3}{2}}}$

Step 3. Evaluating limk→∞ akbk

$The ratio \lim_{k \to \infty} \frac{a_{k}}{b_{k}} is given by : \lim_{k \to \infty} \frac{a_{k}}{b_{k}} = \lim_{k \to \infty} \frac{\frac{k^{\frac{3}{2}} - k - 1}{5 k^{3} + 3}}{\frac{1}{k^{\frac{3}{2}}}} (Substitution) = \lim_{k \to \infty} \frac{k^{\frac{3}{2}} (k^{\frac{3}{2}} - k - 1)}{5 k^{3} + 3} (Simplify) = \lim_{k \to \infty} \frac{k^{3} (1 - \frac{1}{k^{\frac{1}{2}}} - \frac{1}{k^{\frac{3}{2}}})}{k^{3} (5 + \frac{3}{k^{3}})} = \lim_{k \to \infty} \frac{k^{3} (1 - \frac{1}{k^{\frac{1}{2}}} - \frac{1}{k^{\frac{3}{2}}})}{k^{3} (5 + \frac{3}{k^{3}})} (cancel out common factor) = \frac{1}{5}$

Step 4. Result

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In Example 1 we used the comparison test to show that the series $\sum_{k = 1}^{\infty} \frac{k^{\frac{3}{2}} - k - 1}{5 k^{3} + 3}$ converges. Use the limit comparison test to prove the same result.

Short Answer

Step by step solution

Step 1. Given information is:

Step 2. Finding the term ∑k=1∞ bk

Step 3. Evaluating limk→∞ akbk

Step 4. Result

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In Example 1 we used the comparison test to show that the series∑k=1∞k32-k-15k3+3 converges. Use the limit comparison test to prove the same result.

Short Answer

Step by step solution

Step 1. Given information is:

Step 2. Finding the term ∑k=1∞ bk

Step 3. Evaluating limk→∞ akbk

Step 4. Result

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

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Study anywhere. Anytime. Across all devices.

In Example 1 we used the comparison test to show that the series $\sum_{k = 1}^{\infty} \frac{k^{\frac{3}{2}} - k - 1}{5 k^{3} + 3}$ converges. Use the limit comparison test to prove the same result.