Chapter 9.4, Problem 42E

Solution

To state: Whether the statement is true or false. If ${\displaystyle \sum _{n=1}^{\infty }{a}_n}\text{and}{\displaystyle \sum _{n=1}^{\infty }{b}_n}$ both diverge, then ${\displaystyle \sum _{n=1}^{\infty }\left(a_n+b_n\right)}$ diverges.

The statement is false.

Given information:

It is given that if ${\displaystyle \sum _{n=1}^{\infty }{a}_n}\text{and}{\displaystyle \sum _{n=1}^{\infty }{b}_n}$ both diverge, then ${\displaystyle \sum _{n=1}^{\infty }\left(a_n+b_n\right)}$ diverges.

Explanation:

A geometric series ${\displaystyle {\sum }_{k=1}^{\infty }a\cdot r^{k-1}}$ , converges if and only if $\left|r\right|<1$ .If it does converge, the sum: $S=\frac{a_1}{1-r}$ .

Let’s consider a geometric series,

  ${\displaystyle \sum _{n=1}^{\infty }{a}_n}=\frac{1}{48}+\frac{1}{16}+\frac{3}{16}+\frac{9}{16}+\cdots$

The common ratio ( r ) is:

  $\begin{array}{l}r=\frac{\frac{1}{16}}{\frac{1}{48}},r=3\\ r=\frac{\frac{3}{16}}{\frac{1}{16}},r=3\end{array}$

For the given series the value of $a_1$ is $\frac{1}{48}$ and $r\text{is}3$ . As the common ratio $\left|r\right|>1$ , then the given series diverges. It means the sum of the series does not exist even if it consist all positive numbers.

Similarly, let’s consider another geometric series,

  ${\displaystyle \sum _{n=1}^{\infty }{b}_n}=-\frac{1}{48}-\frac{1}{16}-\frac{3}{16}-\frac{9}{16}-\cdots$

The common ratio ( r ) is:

  $\begin{array}{l}r=\frac{-\frac{1}{16}}{\frac{1}{48}},r=3\\ r=\frac{-\frac{3}{16}}{-\frac{1}{16}},r=3\end{array}$

For the given series the value of $a_1$ is $-\frac{1}{48}$ and $r\text{is}3$ . As the common ratio $\left|r\right|>1$ , then the given series diverges. It means the sum of the series does not exist even if it consist all positive numbers.

Now find the sum of ${\displaystyle \sum _{n=1}^{\infty }\left(a_n+b_n\right)}$ ,

  $\begin{array}{c}{\displaystyle \sum _{n=1}^{\infty }\left(a_n+b_n\right)}=\left(\frac{1}{48}+\frac{1}{16}+\frac{3}{16}+\frac{9}{16}+\cdots \right)+\left(-\frac{1}{48}-\frac{1}{16}-\frac{3}{16}-\frac{9}{16}-\cdots \right)\\ =0\end{array}$

Therefore, series converges because it has sum.

Then the statement “If ${\displaystyle \sum _{n=1}^{\infty }{a}_n}\text{and}{\displaystyle \sum _{n=1}^{\infty }{b}_n}$ both diverge, then ${\displaystyle \sum _{n=1}^{\infty }\left(a_n+b_n\right)}$ diverges” is false.