For example: We know that $$\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$$
and we can do some substitutions to this like substitute the x for 2x, then we get
$$\frac{1}{1-2x}=\sum_{n=0}^{\infty}(2x)^n$$
I can understand this step sice we can rewrite the term like $2^n(x-0)^n$, it fits the general form of power series, which is $c_n(x-a)^n$.
However, if we substitude it by $x^2$, the form is $x^{2n}$, this do not fit the general form at all but still it work, what really confusing is someone told me that we can substitute by a $x^n$ where n is a possive intgeral number, but like $\sqrt{x}, 1/x$, we can not substitute, why? Can anyone help me figure it out?
First, let us make clear that we are talking about real funcions of a real variable on the left side and series wih real coefficients on the right side. Second, note that there are two kinds of series (i) formal series where you don't concern yourself with convergence at all but just regad a series as a function $f: \mathbb N \to \mathbb R$ and (ii) series wih an actual number for $x$, where you care about whether or not $$\lim_{n \to \infty}\sum_{n=i}^n(g(x))^i\text { exists.}$$ Let's assume that you are concerned eith the latter kind of series. So you can start with he formula $$\frac{1}{1-y}=\sum_{n=1}^{\infty}y^n \text{ for }-1<y<1$$ and a function $g: \text{ subset of }\mathbb R\to \mathbb R$ and replace $y$ by any $x$ such that $g(x)$ is defined and $-1<g(x)<1$. Depending on the function $g$ and the value of $x$, what you get on the right side may or may not be a power series.