The form of a series I have a problem with series.
Consider a series
$$\sum_{k=1}^\infty a_k.$$ Can I rewrite it as $$\sum_{k=1}^\infty a_k=\sum_{k=1}^\infty a_{2k-1}+\sum_{k=1}^\infty a_{2k}\  ?$$  Or more generally, can I rewrite it as 
$$\sum_{k=1}^\infty a_k=\sum_{k=1}^\infty a_{\sigma_k}+\sum_{k=1}^\infty a_{\delta_k},$$ where $\sigma,\delta$ is the selection of $k$?
I'm considering the proof
$$S_{2n}=\sum_{k=1}^n a_{2k-1}+\sum_{k=1}^n a_{2k},$$ then get the limit when $n\to \infty$, but I don't think it is right.
 A: As you've already noticed, the series must be absolutely convergent for this to be true. To see even better that all hell breaks loose when the series is only conditionally convergent, consider Riemann series theorem: if $\sum a_n$ is a conditionally convergent series, then for every real number $x$, (or even $x=\pm\infty$) there is a permutation $\sigma \in \mathfrak{S}(\mathbb{N})$ such that $\sum_{n=0}^\infty a_{\sigma(n)} = x$.
A: Yes ,it can write as that,because $\sum_{k=1}^{2n}a_k=\sum_{k=1}^n a_{2k-1}+\sum_{k=1}^n a_{2k}$, when $n \to \infty$,
then we have
$\sum_{k=1}^\infty a_k=\sum_{k=1}^\infty a_{2k-1}+\sum_{k=1}^\infty a_{2k}$, if when $n \to \infty$, $\sum_{k=1}^\infty a_{2k-1}$ and $\sum_{k=1}^\infty a_{2k}$ exist.
note: If $\lim_{n \to \infty}\sum_{k=1}^n a_k$ converges, then $\sum_{k=1}^\infty a_k=\lim_{n \to \infty}\sum_{k=1}^n a_k$,this is definition of series. But if when $n \to \infty$, $\sum_{k=1}^\infty a_{2k-1}$ and $\sum_{k=1}^\infty a_{2k}$ are not all exist. You can not write as that.
A: In general: If you have two convergent series:
$$\sum_{k=1}^{\infty} b_k \quad\text{and}\quad \sum_{k=1}^{\infty} c_k.
$$
Then the series 
$$
\sum_{k=1}^{\infty}(b_k + c_k)
$$
is also convergent. So for example if $b_k$ and $c_k$ are given by $b_k = a_{2k-1}$ and $c_k = a{2k}$, then indeed you get a convergent series
$$\sum_{k=1}^{\infty} a_{2k-1} + \sum_{k=1}^{\infty} a_{2k} = \sum_{k=1}^{\infty} a_{2k-1} + a_{2k} = \sum_{k=1}^{\infty} a_k.
$$
Note that for this to be true, you do not need anything to be absolutely convergent.
Now if you rearrange the natural numbers $\mathbb{N}$ so that as $k$ varies in $\mathbb{N}$ then $\delta_k$ also varies through $\mathbb{N}$ (i.e. $\delta : k \mapsto \delta_k$ is a bijection, then in general it is not true that
$$\sum_{k=1}^\infty a_k = \sum_{k=1}^{\infty} a_{\delta_k}
$$
even if the series is convergent. But if the series is absolutely convergent, then this is true.
