Chapter 8.6, Problem 65E

A fallacy Explain the fallacy in the following argument.

Let $x=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots$ and $y=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\cdots .$ It follows that 2y = x + y, which implies that x = y. On the other hand,

$x-y=\underset{>0}{\underset{︸}{\left(1-\frac{1}{2}\right)}}+\underset{>0}{\underset{︸}{\left(\frac{1}{3}-\frac{1}{4}\right)}}+\underset{>0}{\underset{︸}{\left(\frac{1}{5}-\frac{1}{6}\right)}}+\cdots >0$

is a sum of positive terms, so x > y. Therefore, we have shown that x = y and x > y.