Chapter 7: Q. 31 (page 631)

Use any convergence test from this section or the previous section to determine whether the series in Exercises $31 - 48$ converge or diverge. Explain how the series meets the hypotheses of the test you select.
$\sum_{k = 1}^{\infty} k^{\frac{- 1}{2}}$ .

Short Answer

Expert verified

The series $\sum_{k = 1}^{\infty} k^{\frac{- 1}{2}}$ is divergent.

Step by step solution

Step 1. Given information

$\sum_{k = 1}^{\infty} k^{\frac{- 1}{2}}$ .

Step 2. The comparison test states that ∑k=1∞ ak and ∑k=1∞ bk be two terms with positive terms then,

If $\lim_{k \to \infty} \frac{a_{k}}{b_{k}} = L$ , where $L$ is any positive real number.
If $\lim_{k \to \infty} \frac{a_{k}}{b_{k}} = 0$ and role="math" localid="1649222738593" $\sum_{k = 1}^{\infty} b_{k}$ converges, then $\sum_{k = 1}^{\infty} a_{k}$ also converges.
If $\lim_{k \to \infty} \frac{a_{k}}{b_{k}} = \infty$ , and $\sum_{k = 1}^{\infty} b_{k}$ diverges, then $\sum_{k = 1}^{\infty} a_{k}$ also diverges.

Step 3. The terms of the series ∑k=1∞ 1k12 are positive.

Find $\sum_{k = 1}^{\infty} b_{k}$ for the given series.

$\sum_{k = 1}^{\infty} b_{k} = \sum_{k = 1}^{\infty} \frac{1}{k^{\frac{1}{2}}}$

Next find $\lim_{k \to \infty} \frac{a_{k}}{b_{k}}$ for the given series.

$\lim_{k \to \infty} \frac{a_{k}}{b_{k}} = \lim_{k \to \infty} \frac{\frac{1}{k^{\frac{1}{2}}}}{\frac{1}{k^{\frac{1}{2}}}} = \lim_{k \to \infty} 1 = 1$

Step 4. From the obtained values,

The value of $\lim_{k \to \infty} \frac{a_{k}}{b_{k}} = 1$ which is a finite non zero number.

The series $\sum_{k = 1}^{\infty} b_{k} = \sum_{k = 1}^{\infty} \frac{1}{k^{\frac{1}{2}}}$ is divergent by $p -$ series test.

Therefore, the series $\sum_{k = 1}^{\infty} a_{k}$ is also divergent.

Hence, the given series is divergent.

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Use any convergence test from this section or the previous section to determine whether the series in Exercises $31 - 48$ converge or diverge. Explain how the series meets the hypotheses of the test you select.
$\sum_{k = 1}^{\infty} k^{\frac{- 1}{2}}$ .

Short Answer

Step by step solution

Step 1. Given information

Step 2. The comparison test states that ∑k=1∞ ak and ∑k=1∞ bk be two terms with positive terms then,

Step 3. The terms of the series ∑k=1∞ 1k12 are positive.

Step 4. From the obtained values,

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Use any convergence test from this section or the previous section to determine whether the series in Exercises 31-48converge or diverge. Explain how the series meets the hypotheses of the test you select.∑k=1∞k-12.

Short Answer

Step by step solution

Step 1. Given information

Step 2. The comparison test states that ∑k=1∞ ak and ∑k=1∞ bk be two terms with positive terms then,

Step 3. The terms of the series ∑k=1∞ 1k12 are positive.

Step 4. From the obtained values,

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

Theoretical and Mathematical Physics

Decision Maths

Logic and Functions

Mechanics Maths

Pure Maths

Study anywhere. Anytime. Across all devices.

Use any convergence test from this section or the previous section to determine whether the series in Exercises $31 - 48$ converge or diverge. Explain how the series meets the hypotheses of the test you select.
$\sum_{k = 1}^{\infty} k^{\frac{- 1}{2}}$ .