Absolute convergence implies $\sum _{n=1}^{\infty} a_n= \sum _{n=1} ^{\infty} a_{2n}+ \sum _{n=1} ^{\infty} a_{2n-1}$

Question

Let $\sum _{n=1}^{\infty} a_n$ be a series of complex numbers converging absolutely. Then I know that by Riemann's rearrangement theorem, the value of the sum is independent of rearrangements. But I am wondering that $\sum _{n=1}^{\infty} a_n= \sum _{n=1} ^{\infty} a_{2n}+ \sum _{n=1} ^{\infty} a_{2n-1}$ holds (is this also a rearrangement?)

Answer 1

\begin{align} \sum_{n=1}^\infty a_n &= a_1 + a_2 + a_3 + a_4 + \ldots \\ &= (a_1 + a_2) + (a_3 + a_4) + \ldots \\ &= \sum_{n=1}^\infty (a_{2n-1}+a_{2n}) \\ &= \sum_{n=1}^\infty a_{2n-1}+\sum_{n=1}^\infty a_{2n} \\ \end{align}

Let the partial sum be $s_n = \sum_{k=1}^n a_k$. Grouping the terms means to look only at the even partial sums $(s_{2n})$, which is a subsequence of the original convergent sequence of partial sum, and must therefore also converge (to the same limit).

Since the series $\sum_{k=1}^\infty a_k$ converges absolutely, the series $\sum_{k=1}^\infty a_{2k-1}$ and $\sum_{k=1}^\infty a_{2k}$ converge absolutely as well (and therefore converge). Consider the partial sums $x_n = \sum_{k=1}^n |a_{2k-1}|$ and $y_n = \sum_{k=1}^n |a_{2k}|$. Since $(a_n)$ is absolutely convergent, $(x_n)$ and $(y_n)$ are bounded above. They are also increasing, since $x_{n+1} - x_n = |a_{2n+1}| \geq 0$ and $y_{n+1} - y_n = |a_{2n+2}| \geq 0$. By the monotone convergence theorem, the sequences $(x_n)$ and $(y_n)$ will converge.

The last part is an application of the limit law (as mentioned by conditionalMethod) $\lim_{n\to \infty} (x_n + y_n) = \lim_{n\to \infty} x_n + \lim_{n\to \infty} y_n$, if the limits of the right hand side exist.