Showing that $\sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} a_{2n} + \sum_{n=1}^{\infty} a_{2n+1}$

Question

This is what I am trying to prove:

If $\sum_{n=1}^{\infty} a_n$ is an absolutely convergent series with complex terms then show that $\sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} a_{2n} + \sum_{n=1}^{\infty} a_{2n-1}$.

---

Here's my attempt:

To prove this, I am trying to use the fact that absolute convergence of the series implies that the terms can be rearranged in any fashion desired and we can possibly group terms to get the desired consequence. However, it is not immediate to me which bijection $\sigma : \mathbb N \to \mathbb N$ to use so that \begin{align} \sum_{n=0}^{\infty}a_n =\sum_{n=0}^{\infty} a_{\sigma(n)} \end{align}

Hints are appreciated!

---

EDIT: Here's an alternative approach that I found:

Clearly, $\sum_{n=1}^{\infty} |a_{2n}| \le \sum_{n=1}^{\infty} |a_n|$. Hence, $\sum_{n=1}^{\infty} a_{2n}$ is absolutely convergent (and hence convergent!). The same can be said about $\sum_{n=1}^{\infty} a_{2n-1}$. Hence, we have that $\sum_{n=1}^{\infty} a_{2n} + \sum_{n=1}^{\infty} a_{2n-1}= \sum_{n=1}^{\infty} (a_{2n} + a_{2n-1}) = \sum_{n=1}^{\infty} a_n$. Note that the first equality holds because the series of odd terms and even terms converge (as shown) and the last equality holds because grouping of terms does not affect convergence.

Answer 1

Let $b_n=(-1)^na_n$. Then $\sum b_n$ is also absolutely convergent. We have $$\frac12\sum a_n +\frac12\sum b_n=\sum\frac{a_n+b_n}2=\sum a_{2n}$$ where the last step is to drop the zero summands at odd indices. Likewise, $$\frac12\sum a_n-\frac12\sum b_n=\sum a_{2n-1}.$$ By adding these equations, the result follows.

Answer 2

The idea is not bad but you have to be more "sneaky". Let $(\sigma_k:\mathbb{N}\to \mathbb{N})_{k\in \mathbb{N}}$ be the sequence of permutations: $$\sigma_k(n)=\begin{cases}2n \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n=1,...,k \\ 2(n-k)-1 \ \ \ \ \ n=k+1,...,2k \\ n \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{otherwise}\end{cases}$$ So: $$\sum_{n=1}^{\infty} a_n=\sum_{n=1}^{\infty} a_{\sigma_k(n)}$$ $$\lim_{s\to \infty}\sum_{n=1}^{s} a_n=\lim_{s\to \infty} \sum_{n=1}^{s} a_{\sigma_k(n)}$$ $$\lim_{s\to \infty}\sum_{n=1}^{s} a_n=\lim_{s\to \infty} \sum_{n=1}^{k} a_{2n}+\sum_{n=k+1}^{2k} a_{2(n-k)-1}+\sum_{n=2k+1}^{s} a_{2(n-k)-1}$$ $$\sum_{n=1}^{\infty} a_n=\sum_{n=1}^{k} a_{2n}+\sum_{n=k+1}^{2k} a_{2(n-k)-1}+\sum_{n=2k+1}^{\infty} a_{n}$$ By taking $k\to\infty$ the last sum vanishes because it's the rest of a convergent series: $$\sum_{n=1}^{\infty} a_n=\lim_{k\to \infty}\sum_{n=1}^{k} a_{2n}+\sum_{n=k+1}^{2k} a_{2(n-k)-1}$$ The second sum can be rewritten with a new index as: $$\sum_{n=1}^{\infty} a_n=\lim_{k\to \infty}\sum_{n=1}^{k} a_{2n}+\sum_{n=1}^{k} a_{2n-1}$$ $$\sum_{n=1}^{\infty} a_n=\sum_{n=1}^{\infty} a_{2n}+\sum_{n=1}^{\infty} a_{2n-1}$$