Chapter 7: Q.2C) (page 631)

A p series other than $\sum_{k = 1}^{\infty} \frac{1}{k^{2}}$ you could use with comparison test to show that the series $\sum_{k = 1}^{\infty} \frac{k - 1}{k^{3} + k + 1}$ converges.

Short Answer

Expert verified

It is convergent

Step by step solution

Step 1

Consider the series $\sum_{k = 1}^{\infty} \frac{k - 1}{k^{3} + k + 1}$

To determine p series that used to show that $\sum_{k = 1}^{\infty} \frac{k - 1}{k^{3} + k + 1}$ is convergent

The terms of series $\sum_{k = 1}^{\infty} \frac{k - 1}{k^{3} + k + 1}$ are positive.

The series $\sum_{k = 1}^{\infty} b_{k}$ for the series $\sum_{k = 1}^{\infty} \frac{k - 1}{k^{3} + k + 1}$ is

$\sum_{k = 1}^{\infty} b_{k} = \sum_{k = 1}^{\infty} \frac{1}{k^{3 / 2}}$

c) Step 2

The ratio $\lim_{k \to \infty} \frac{a_{k}}{b_{k}}$ is given

$\lim_{k \to \infty} \frac{a_{k}}{b_{k}} = \lim_{k \to \infty} \frac{\frac{k - 1}{k^{3} + k + 1}}{\frac{1}{k^{3 / 2}}} = \lim_{k \to \infty} \frac{k^{3 / 2} (k - 1)}{k^{3} + k + 1} = \lim_{k \to \infty} \frac{k^{5 / 2} (1 + \frac{1}{k})}{k^{3} (1 + \frac{1}{k^{2}} + \frac{1}{k^{3}})} = \lim_{k \to \infty} \frac{(1 + \frac{1}{k})}{k^{1 / 2} (1 + \frac{1}{k^{2}} + \frac{1}{k^{3}})} = 0$

c) Step 3

The value 0f $\lim_{k \to \infty} \frac{a_{k}}{b_{k}} = 0$

The series $\sum_{k = 1}^{\infty} b_{k} = \sum_{k = 1}^{\infty} \frac{1}{k^{3 / 2}}$ is convergent by the p-series test

Then $\sum_{k = 1}^{\infty} a_{k}$ is also convergent

Then the series $\sum_{k = 1}^{\infty} \frac{k - 1}{k^{3} + k + 1}$ is convergent and the p series is $\sum_{k = 1}^{\infty} b_{k} = \sum_{k = 1}^{\infty} \frac{1}{k^{3 / 2}}$

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A p series other than $\sum_{k = 1}^{\infty} \frac{1}{k^{2}}$ you could use with comparison test to show that the series $\sum_{k = 1}^{\infty} \frac{k - 1}{k^{3} + k + 1}$ converges.

Short Answer

Step by step solution

Step 1

c) Step 2

c) Step 3

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A p series other than ∑k=1∞1k2you could use with comparison test to show that the series ∑k=1∞k-1k3+k+1converges.

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A p series other than $\sum_{k = 1}^{\infty} \frac{1}{k^{2}}$ you could use with comparison test to show that the series $\sum_{k = 1}^{\infty} \frac{k - 1}{k^{3} + k + 1}$ converges.