Chapter 9.6, Problem 53ES

Exercise 51 illustrates that one of the nuances of “conditional� convergence is that the sum of a series that converges conditionally depends on the order that the terms of the series are summed. Absolutely convergent series are more depend- able, however. It can be proved that any series that is con- structed from an absolutely convergent series by rearranging the terms will also be absolutely convergent and has the same sum as the original series. Use this fact together with parts (a) and (b) of Theorem 9.4.3 in these exercises.

Use the series for ${\pi }^2/6$ given in the preceding exercise to show that $\frac{{\pi }^2}{12}=1-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+\cdots$