Rearrangements of Series I'm currently a university student and just completed my my first year. In Analysis I, we discussed how certain infinite series, namely, series which do not converge absolutely, can be modified (i.e. rearranged) so that they converge to different limits.
Now I recall that in sixth form (high school), we proved certain statements by rearranging infinite series. For example, the fact that $$e^{ix}=\cos x+i\sin x$$
we proved by the following reasoning:
$$e^{ix}=\sum_{r=0}^\infty \frac{(ix)^r}{r!}=1+ix-\frac{x^2}{2!}-i\frac{x^3}{3!}+\frac{x^4}{4!}+i\frac{x^5}{5!}-\frac{x^6}{6!}-i\frac{x^7}{7!}+\cdots$$
$$\hspace{1cm}=\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots\right)+i\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots\right)\tag{$*$}$$
$$=\cos x+i\sin x$$
Is this proof valid? Surely we cannot assume that the step in $(*)$, which is clearly a rearrangement of the terms of the original series, still converges to $e^{ix}$, can we? Unless the original series converges absolutely? I'd appreciate any help with the matter.
 A: The series for $e^{ix}$ does converge absolutely, so the re-arrangement is valid.
To see that it converges absolutely, consider the series which is the sum of the absolute values of all those terms: 
$$
\sum_{r=0}^\infty \left| \frac{(ix)^r}{r!} \right| = \sum_{r=0}^\infty  \frac{x^r}{r!} = e^x < \infty
$$
A: When a series (real or complex) converges absolutely its sum is independent of any rearrangement. If the terms go to zero, it converges but does not converge absolutely then you may arrange terms so as to get any finite or infinite number as limit (at least in the real case).
A: When one speaks of rearrangement of an absolutely convergent series, often one means one re-orders the terms so that the set of terms that had indices $0,1,2,3,\ldots$ will have those same indices but which index goes with which term will be different.  So every term will still have only finitely many others preceding it.  Your rearrangement, however, puts infinitely many terms before some of the other terms.  The fact is, that is also valid with absolutely convergent series.  If one has proved that last fact, then there's no problem with your argument.
This series is a power series.  Power series of the form $\displaystyle \sum_{n=0}^\infty a_n x^n$ converge absolutely if $|x|$ is smaller than the radius of convergence of the series, and diverge if $|x|$ is bigger than the radius of convergence.  (What happens when $|x|$ is equal to the radius of convergence depends on which series it is, in a somewhat complicated way.  Sometimes it converges conditionally at some points $x$ on the boundary of the disk of convergence.)
The particular power series you deal with has an infinite radius of convergence, so it converges absolutely if $x$ is any complex number at all.
I've made some assertions that I haven't proved.  If those can be relied upon, then your argument is valid.
$$ \S $$
PS: For some purposes it might be better to think of absolutely convergent series as having terms in no particular order.  Let the series be
$$
\sum\{ a_n : n\in\mathscr S\, \}
$$
where $\mathscr S$ is some set.  (I don't want to write $\displaystyle\sum\{a : a\in \mathscr A\}$ where $\mathscr A$ is a set of numbers, because that would allow every number to occur only once in the sum.)  Then we can define
$$
\sum\{ |a_n| : n\in\mathscr S\, \} = \sup\left\{ \sum \{|a_n| : n\in\mathscr T \} : \mathscr T \text{ is a finite subset of } \mathscr S \right\}.
$$
If that is finite, then the series converges absolutely.  Then we can define
\begin{align}
\mathscr S_0 & = \{ n\in\mathscr S : a_n \ge 0\} \\
\mathscr S_1 & = \{ n\in\mathscr S : a_n < 0\} \\[10pt]
& \sum\{ a_n : n\in\mathscr S\, \} \\
= {} & \sup\left\{ \sum \{a_n : n\in\mathscr T \} : \mathscr T \text{ is a finite subset of } \mathscr S_0 \right\} \\
& {} - \sup\left\{ \sum \{  -a_n : n\in\mathscr T \} : \mathscr T \text{ is a finite subset of } \mathscr S_1 \right\}.
\end{align}
