Can we expand $(1-x)^{-1}$ for $\left| x\right| >1$ I don't know too much  about formal power series, but by using some common sense
I can say that
$$(1-x)^{-1}=1+x+x^2+x^3+x^4\ldots$$  will only hold iff $\left| x\right| <1$.
Does this series hold iff $\left| x\right| >1$?,\
But controversy is seen in "chapter 9 Renormalization" in Ryder's book on QFT at page 341, its as follows,
$$\begin{align}
D_{\mu\nu}'=D_{\mu\nu}-D_{\mu\alpha}\big(k^\alpha k^\beta-g^{\alpha\beta}k^2\big)\Pi(k^2)D_{\beta\nu}
\end{align}$$
and hence putting $D_{\mu\nu}=-g_{\mu\nu}/k^2,$
$$\begin{align}
D'_{\mu\nu}(k)&=\frac{1}{k^2[1+\Pi(k^2)]}\Bigg(-g_{\mu\nu}-\frac{k_\mu k_\nu}{k^2}\Pi(k^2)\Bigg)\\&=\frac{-g_{\mu\nu}}{k^2[1+\Pi(k^2)]} + \text{gauge terms}.  
\end{align}\tag{9.122}$$...
I tried a lot to attain his final formulae, Finally, I find that,
if $\Pi(k^2)$ is possible to expand as a power series, then I get the desired result,
So I expand
$$(1+\Pi(k^2))^{-1}=1+\Pi(k^2)+(\Pi(k^2))^2+\ldots$$ then I get the result given by Ryder by simple algebraic expansion as follows,
putting $D_{\mu\nu}=-\frac{g_{\mu\nu}}{k^2}$(Feynman's propagator)    gives
$$\begin{align}
D'_{\mu\nu}(k)=\frac{-g_{\mu\nu}}{k^2}-\Bigg(\frac{k_\mu k^\nu}{k^2}-g_{\mu\nu}\Bigg)\frac{\Pi(k^2)}{k^2}
=\frac{1}{k^2}\big(1-\Pi(k^2)\big)\Bigg[-g_{\mu\nu}-\frac{k_\mu k^\nu\Pi(k^2)}{1-\Pi(k^2)}\Bigg].\end{align}$$
Iff $\Pi(k^2)<<1$
$$\begin{align}
D'_{\mu\nu}(k)&=\frac{1}{k^2[1+\Pi(k^2)]}\Bigg(-g_{\mu\nu}-\frac{k_\mu k_\nu}{k^2}\Pi(k^2)\Bigg)\\&=\frac{-g_{\mu\nu}}{k^2[1+\Pi(k^2)]} + \text{gauge terms}.   
\end{align}\tag{9.124}$$
But in fact, $\Pi(k^2)>>1$ because $$\Pi(k^2)=\frac{e^2}{6\pi^2\epsilon}+\frac{e^2k^2}{60\pi^2m^2}+\ldots$$, and this $\epsilon$ is very small in fact, its magnitude is almost equal to zero. So we can't use power series in fact, Then how would we say Ryder's derivation is correct??
 A: As a series of real numbers, $(1-x)^{-1}=1+x+x^2+x^3+\ldots$ holds for $|x|\lt 1$. (For $|x|\ge 1$, the RHS diverges.)
As a formal power series (see: Wikipedia), $1-x\in\mathbb R[[x]]$ is invertible, and its inverse is $1+x+x^2+x^3+\ldots$ because $(1-x)(1+x+x^2+\ldots)=1$ in $\mathbb R[[x]]$. Note in this context $x$ is purely a symbol and the multiplication is formal, and does not involve substitution of $x$ with any value.
So, the answer here is that Mathematics has both of those tools for Physics' disposal,  I guess the question is "which one makes physical sense". Apparently using formal power series "works" here - gives results that agree with the experiment.
A: No; since each term on the RHS is increasing by a factor of $x$, if $\left| x\right|>1$, then $$\lim_{n\to \infty}a_n=\pm \infty,$$ so the series diverges.
Another way to think about it is that since $\left| a_n \right|$ is monotonically increasing, each term "cancels out" the effect of the last one and makes the magnitude grow in either the positive or negative direction. Since the series is infinite, this growth in magnitude never stops, so the series diverges.
A: No, with $|x|\ge1$, the general term does not tend to $0$ and the series diverges.
