Rewriting Infinite Sum Are the following equal?
$$\sum_{k = 1}^{\infty}a_k = \sum_{k = 1}^{\infty}(a_{2k} + a_{2k - 1})$$
If I expand the summations they are the same series, so they should be equivalent in all respects (convergence, divergence, etc.), right? I just want to confirm that there is no problem with manipulating an infinite series like this.
 A: It is subtle. Assuming $\sum_{k\geq 1}a_k$ is convergent, you are always allowed to write $$\sum_{k\geq 1}a_k = \sum_{k\geq 1}(a_{2k}+a_{2k-1})$$ but not $$\sum_{k\geq 1}a_k = \sum_{k\geq 1}a_{2k}+\sum_{k\geq 1}a_{2k-1}$$ as testified by $\sum_{n\geq 1}\frac{(-1)^{n-1}}{n}$, that is only conditionally convergent and not absolutely convergent.
That ultimately depends on the fact that
$$ \lim_{n\to +\infty}(a_n+b_n) = \lim_{n\to +\infty}a_n+\lim_{n\to +\infty}b_n $$
as soon as every limit makes sense.
There is also a quite surprising fact, stated by the Riemann series theorem: if you have a series that is conditionally convergent but not absolutely convergent, for any $\zeta\in\mathbb{R}$, you may re-arrange its terms through a permutation $\sigma$ in order that
$$\sum_{n\geq 1}a_{\sigma(n)}=\zeta. $$
A: No, you can't always write $$\sum_{k\geq 1}a_k = \sum_{k\geq 1}(a_{2k}+a_{2k-1})$$ For example, let $a_k = (-1)^k.$ Then the first series diverges, while the second series is just the sum of $0$'s, hence converges to $0.$ It is true that that if the first series converges, then the second series converges to the same value. (But even in this case, the two series are usually not identical as sequences of partial sums.)
A: No, in general not.
the partial sums of the second form are only a subsequence of the partial sums of the first.
However, with the further assumption that the sequence of terms $(a_k)_k$ is a zero sequence, the two forms are equivalent.
