Chapter 8: Problem 65
A fallacy Explain the fallacy in the following argument. Let \(x=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots\) and \(y=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\cdots \cdot\) It follows that \(2 y=x+y\) which implies that \(x=y .\) On the other hand, $$ x-y=\underbrace{\left(1-\frac{1}{2}\right)}_{>0}+\underbrace{\left(\frac{1}{3}-\frac{1}{4}\right)}_{>0}+\underbrace{\left(\frac{1}{5}-\frac{1}{6}\right)}_{>0}+\cdots>0 $$ is a sum of positive terms, so \(x>y .\) Therefore, we have shown that \(x=y\) and \(x>y\)
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