The order of summation of $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} i^n$. This is a follow-up to my previous problem. I think it will be clearer if I ask this as a stand-alone question. I apologise if I come up a bit hectic but I'm not entirely sure what my doubts are exactly.
Basically, at 16:30 here we change the order of summation of $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} i^n$, separating this into series of real and imaginary parts. My question is - how do we know that it doesn't change the sum of the original series if it's not absolutely convergent? Is there a result that states this? The only theorem I've come across is:
The necessary and sufficient condition that the series of complex numbers
$$\alpha_1 + \alpha_2 + \alpha_3 + \dots + \alpha_n + \dots$$
converges to a limit $\alpha = a + bi$ is that
$$L_{n = \infty} A_n = a, \; L_{n = \infty} B_n = b.$$
However, I don't see how that helps. It seems to me like we just take it for granted that $1 - \frac{1}{3} + \frac{1}{5} - \dots$ (after rearrangement) sums up to $\frac{\pi}{4}$ (before rearrangement).
 A: If $\sum a_n$, $\sum b_n$ and $\sum a_n + b_n$ are all convergent, you have
$$\sum a_n + b_n = \sum a_n + \sum b_n$$
This has nothing to do with a rearrangement of the terms, there is no reordering here. This just uses the fact that when all the limits exist, one has
$$\lim (u_n + v_n) = \lim u_n + \lim v_n$$
A: Let $z_n=a_n+ib_n$ be a sequence of complex numbers, where $a_n,b_n\in\mathbb{R}$ are the real and imaginary parts. In general, the (complex) series $\sum z_n$ is convergent if and only if the two (real) series $\sum a_n$ and $\sum b_n$ are both convergent.
Basically, this is because
$|a+bi|\le |a|+|b|\le 2|a+bi|$ and thus
$$
|\sum_{n=1}^N z_n-L|
\le |\sum_{n=1}^N a_n-\textrm{Re}(L)|+|\sum_{n=1}^N b_n-\textrm{Im}(L)|
\le 2|\sum_{n=1}^N z_n-L|
$$
If we know that $\sum z_n$ is convergent (and thus both the series of the real and imaginary parts convergent), the partial sum identity
$$
\sum_{n=1}^N z_n = \sum_{n=1}^N a_n+i\sum_{n=1}^N b_n
$$
implies that
$$
\sum_{n=1}^\infty z_n = \sum_{n=1}^\infty a_n+i\sum_{n=1}^\infty b_n
$$
Note that in the discussion above, we don't use anything about $\sum |z_n|$, i.e., we don't need to know whether $\sum z_n$ is absolutely convergent or not.
