Chapter 9.5, Problem 34E

a.

To determine

The objective is to explain how to rearrange the terms of the given series to form a divergent series.

Answer to Problem 34E

Begin by adding positive terms until the partial sum is greater than 1. Then add negative terms until the result is -2. Next add positive terms until the partial sum is greater than 3 then add negative terms unit the sum is less than -4 and continue like this manner indefinitely, so that the sequence of partial sums swings arbitrarily far in both directions.

Explanation of Solution

Given information:

Series: ${\displaystyle \sum _{n=2}^{\infty }{(-1)}^{n+1}}\frac{1}{n\ln n}$ , an alternating series.

Concept used:

Rearrangements of conditionally convergent series:

If ${\displaystyle \sum a_n}$ converges conditionally, then the terms can be rearranged to form a divergent series.

Given any real number, the terms can be rearranged to form a series that converges to that number.

Here,

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

The positive terms diverge to $\infty$ while the negative terms diverge to $-\infty$ .

For rearrangement of the series, start by adding positive terms until the partial sum is greater than1. Then add negative terms until the result is -2.

Then again add positive terms until the partial sum is greater than 3 and again add negative terms unit the sum is less than -4.

Continue like this manner indefinitely, so that the sequence of partial sums swings arbitrarily far in both directions and thus the series diverges.

b.

To determine

The objective is to explain how to rearrange the terms of the series from the specified exercise to form a series the converges to 4.

Answer to Problem 34E

Start by adding positive terms until the partial sum is greater than 4. Then add negative terms until the partial sum is less than 4. Then, again add positive terms until the sum is greater than 4 and continue in this manner indefinitely, always closing in on 4.

Explanation of Solution

Given information:

Series: ${\displaystyle \sum _{n=2}^{\infty }{(-1)}^{n+1}}\frac{1}{n\ln n}$ , an alternating series.

Concept used:

Rearrangements of conditionally convergent series:

If ${\displaystyle \sum a_n}$ converges conditionally, then the terms can be rearranged to form a divergent series.

Given any real number, the terms can be rearranged to form a series that converges to that number.

Here,

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

Initially start by adding positive terms until the partial sum is greater than 4.

Then add negative terms until the partial sum is less than 4.

Then, again add positive terms until the sum is greater than 4 and so on.

Continue in this manner indefinitely, always closing in on 4.

Since the positive and negative terms of the original series both approach zero, the amount by which the partial sums exceed or fall short of 4 approaches zero.