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

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.

• Note that this is just the odd numbered terms added to the even terms. This means all you have to do is rearrange the sum on the right hand side and it will be the sum on the left. Jan 8 at 14:46
• $\sum^N_{j=1}a_n=\sum^{\lfloor N/2\rfloor}_{j=1}a_{2j}+\sum^{\lfloor N/2\rfloor}_{j=0}a_{2j+1}$. For general rearrangements and subsettings, see Apostol, T. Mathematical Analysis, 2nd edition, Addison Wesley, 1974 pp. 187, 196-199. Jan 8 at 15:06
• @OliverDíaz This looks but this does not make use of the rearrangement theorem. The absolute convergence guarantees that the sums in the right hand converge. Jan 8 at 16:30
• Do the map from right to left so that you're mapping the even termed sum $a_{2n} \rightarrow a_{2n}$ and odd termed sum $a_{2n+1} \rightarrow a_{2n+1}$ which is clearly injective and surjective on the full sum. Since it's a bijection it's invertible and that inverse provides the required rearrangement. Jan 8 at 16:55
• @CyclotomicField I am unable to make sense of your notation. Do you mind taking a look at attempt that I have added? Jan 8 at 18:18

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.
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: $$\sum_{n=1}^{\infty} a_n=\lim_{k\to \infty}\sum_{n=1}^{k} a_{2n}+\sum_{n=1}^{k} a_{2n-1}$$ $$\sum_{n=1}^{\infty} a_n=\sum_{n=1}^{\infty} a_{2n}+\sum_{n=1}^{\infty} a_{2n-1}$$
• This looks good but it does not make best use of the "rearrangement theorem" as the the bijection $\sigma _k$ just permutes just finitely many terms and the others stay put. Jan 8 at 16:28
• You can't construct a permutation that puts all of the even numbers first and then the odd numbers. This problem is equivalent to finding an order-preserving bijection between $\mathbb{N}$ and $\mathbb{Z}$. Jan 8 at 16:38