Question 4.T.5: If ∑xn is absolutely convergent, then all its rearrangements...
If \sum{x_{n}} be absolutely convergent, then all its rearrangements also converge absolutely to the same limit.
Let \sum{x_{n}} be absolutely convergent, and suppose \sum{y_{n}} is a rearrangement of \sum{x_{n}}. Define
S_{n} = \sum\limits_{k=1}^{n}{x_{n}}, S = \lim S_{n}, T_{n} = \sum\limits_{k=1}^{n}{y_{k}},and let ε > 0. By the absolute convergence of \sum{x_{n}}, there is an integer N such that
|S − S_{N} | = \sum\limits_{k=N+1}^{∞}{|x_{k}|} = |x_{N+1}| + |x_{N+2}| + · · · < \frac{ε}{2}.Now we choose the integer M so that all the terms x_{1}, x_{2}, · · · , x_{N} appear in the first M terms of the rearranged series, that is, within the finite sequence (y_{1}, y_{2}, · · · , y_{M} ). Hence these terms do not contribute to the difference T_{m} − S_{N} , where m ≥ M, and we obtain
m ≥ M ⇒ |T_{m} − S_{N} | ≤ |x_{N+1}| + |x_{N+2}| + · · · < \frac{ε}{2}⇒ |S − T_{m}| ≤ |S − S_{N} | + |S_{N} − T_{m}| < ε.
Thus lim T_{n} = S.
To show that the rearranged series \sum{y_{n}} is absolutely convergent, we note that \sum{|y_{n}|} is a rearrangement of the absolutely convergent series \sum{|x_{n}|} and, by the argument above, converges to the same limit.