What are the rules for changing indexes and powers in a series? My TA told me that when I want to change indexes in a series, I can make a substitution just like when I'm doing integral by substitution. But I think I have encountered some cases where this goes wrong. So can anybody tell me what substitutions are allowed if I want to rewrite a series?

As an example, if I have a geometric series:
$$
\sum_{n=2}^{\infty} z^{n-2}=\sum_{k=0}^{\infty} z^{k}
$$
I can change the index and power by substituting k=n-2 and change the limits correspondingly. This substitution is valid.
But If you instead consider a power series of the form:
$$
\sum_{n=0}^{\infty} c_{n} x^{2 n+1} \neq \sum_{k=1}^{\infty} c_{\frac{k-1}{2}} x^{k}
$$
I believe it's generally untrue, that I can just substitute $k=2n+1$, so $n=(k-1)/2$.
A specific case where I think this is wrong is this series, which I tried to rewrite:
$$\sum_{n=0}^{\infty}(-1)^{n} \frac{x^{2 n}}{(2 n) !}= \sum_{k=0}^{\infty}(-1)^{k / 2} \frac{x^{k}}{k !}=\sum_{k=0}^{\infty} \frac{i^{k} x^{k}}{k !}$$
With k=2n.
But I'm pretty certain that the equal sign is wrong (since the first is the cosine series, and the second is some complex series), and thus what my TA told me must be wrong?

The reason why I wanted to learn how to rewrite the series, is that I often have formulas, that only work for specific expressions. For example the sum of a geometric series, or how to find the radius of convergence:
Theorem: Let $\sum_{n=0}^{\infty} a_{n} x^{n}$ Be a power series. The limit
$$
R=\lim _{n \rightarrow \infty}\left|\frac{a_{n}}{a_{n+1}}\right|
$$ is the radius of convergence for the series.
 A: The issue lies in the meaning of the symbol $\sum_{k=a}^\infty$. It means that, in the sum, $k$ takes each of the values $a,a+1,a+2,a+3,...$. If you make a transformation like $n=2k+1$, say with the initial index value  $a=1$, the symbolism doesn't work any more. The first index value can be changed, correctly, to $n=3$, but then already the next index value is wrong: It should be $n=5$, but the summation symbol $\sum_{n=3}^\infty$ makes the next index value $4$ (for which there is no corresponding $k$ value).
The rule is: the summation variable can be changed by a simple shift (e.g. $n=k-1$ or $j=k+2$), when only the first (and perhaps the  last) index values have to be taken care of. Any transformation involving scaling, if possible at all, needs special treatment—for example, dividing the series into sums of odd-indexed and even-indexed terms.
A: Consider the power series$$\sum_{n=0}^\infty2^nx^{3n}.\tag1$$I suppose that this is one of those power series to which you would like to apply the formula$$\lim_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right|\tag2$$in order to compute the radius of convergence. But you can't. That is, yes, you can write $(1)$ as a power series of the type$$\sum_{n=0}^\infty a_nx^n.$$The problem is that then$$a_n=\begin{cases}2^{n/3}&\text{ if }3\mid n\\0&\text{ otherwise.}\end{cases}$$and those $0$'s will not let you use formula $(2)$.
But you can solve it in another way: compute$$\left|\lim_{n\to\infty}\frac{2^{n+1}x^{3(n+1)}}{2^nx^n}\right|=\left|\lim_{n\to\infty}{2x^3}\right|,$$which will be $\infty$ if $|x|>\frac1{\sqrt[3]2}$ and $0$ if $|x|<\frac1{\sqrt[3]2}$. Therefore, the radius of convergence is $\frac1{\sqrt[3]2}$.
