Chapter 10.5, Problem 33E

(a)

To determine

To find: The series that is a divergent series.

Answer to Problem 33E

The required series is ${\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\frac{n^2}{1+n}}$ .

Explanation of Solution

Given information: ${\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\frac{1+n}{n^2}}$

Calculation:

  $\Rightarrow {\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\frac{1+n}{n^2}}$

Make the numerator greater than denominator by swapping them to make it divergent series.

  $\therefore {\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\frac{n^2}{1+n}}$

By limit comparison test with divergent series ${\displaystyle \sum _1^{\infty }\frac{1}{n}}$

  $\begin{array}{l}\underset{n\to \infty }{\lim }\frac{\frac{1}{n}}{\frac{n^2}{n+1}}=\underset{n\to \infty }{\lim }\frac{1}{n}\times \frac{n+1}{n^2}\\ =\underset{n\to \infty }{\lim }\frac{n+1}{n^3}\\ =0\end{array}$

The limit is zero; it means this test is inconclusive.

  ${\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\frac{n^2}{1+n}}=\frac{1}{2}-\frac{4}{3}+\frac{9}{4}-\frac{16}{5}+\frac{25}{6}\mathrm{...}$

  $\begin{array}{l}{s}_2\approx -0.833\\ s_3\approx 1.417\\ s_4\approx -1.783\\ s_5\approx 2.383\\ s_6\approx -2.76\end{array}$

Here, $s_n$ is getting further and further away from zero.

Hence, the series is divergent.

(b)

To determine

To find: The series that is a convergent to 4.

Answer to Problem 33E

First add positive terms! if the partial sum is less than 4, add positive terms, and if the partial sum is greater than 4, add negative terms, so that the partial sums are ever growing closer to 4.

Explanation of Solution

Given information: ${\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\frac{1+n}{n^2}}$

Calculation: Since terms of the series is alternate between positive and negative.

Hence, make the partial sum is larger than four by adding positive terms. Then, make the partial sum is lower than four by adding negative terms. Repeat this process indefinitely. At the end $s_n$ is getting closer and closer to four. Hence, the rearranged series must be convergent to four.