How does $\varphi \colon T \mapsto (1+T)^p-1$ chance the radius of convergence for p-adic Laurentseries Let $p\ne 2 $ be a prime. Let $x=\sum_{i\in \mathbb{Z}} a_iT^i$  be a formal Laurentseries with coefficients in a finite extension $L$ of $\mathbb{Q}_p$. I will say $\sum a_iT^i $ converges for $v_p(T)=r\in \mathbb{R}$ when I mean evaluating $\sum a_iT^i $ at some $a\in C_p$ with $v_p(a)=r$ converges. 
1) I am trying to find out how the endomorphism $\varphi$ sending  $T\mapsto (1+T)^p-1$ and fixing $L$ is changing the annulus of convergence of a Laurent serie. Suppose the series $x$ converges for elements of $C_p$ with  $v_p(T)=r$. Then the way $\varphi$ operates on x, by substituting $T$ by $(T+1)^p-1$, we can conclude that $\varphi(x)$ converges for elements of $C_p$ with  $v_p(\varphi(T))=r$. I find 
\begin{align*}
     v_p(\varphi(T))= \begin{cases} v_p(T)+1 &\mbox{if } v_p(T)>\frac{1}{(p-1)}  \\ v_p(T)+v_p(T^{p-1})= p v_p(T) & \mbox{if } v_p(T)<\frac{1}{(p-1)}  \end{cases} 
\end{align*}
but a proplem accures for $v_p(T)=\frac{1}{p-1}$. 
If $T=(-p)^{\frac{1}{p-1}}$ then $T^{p-1}+p=0$, so 
\begin{align*}
v_p(\varphi(T))&= v_p(T)+v_p(p+...+pT^{p-2}+T^{p-1})\\&=v_p(T)+v_p(\tbinom{p}{2}T+...+pT^{p-2})\\&=v_p(T)+v_p(\tbinom{p}{2}T)\\&=1+2v_p(T)=\frac{p+1}{p-1}
\end{align*}
but if $T\ne(-p)^{-p+1}$ with $v_p(T)=\frac{1}{p-1}$ then \begin{align*}v_p(\varphi(T))= v_p(T)+v_p(p+...+pT^{p-2}+ T^{p-1})=v_p(T)+1=  \frac{p}{p-1}.\end{align*}
Suppose now $r=\frac{p}{p-1} $ and $x$ converges only on $r$. Then $\varphi(x)$ would converge for any $T$ with $v_p(T)=\frac{1}{p-1}$,  $T\ne(-p)^{-p+1}$ but not for $T= (-p)^{-p+1}$, which is a contradiction to the fact that in $C_p$ series always converge on circles. 
What is going wrong here? 
 A: Let's look at a simpler example. Take the operator $\psi: T \mapsto 1-T$ and the Laurent series $x = T^{-1}$, which converges for every $v_p(T) = r \in \mathbb{R}$. What is $\psi(x) = \frac{1}{1-T}$ as a Laurent series, and would you claim it gives a convergent series when evaluating $T=a$ with any $v_p(1-a) \in \mathbb{R}$ (in other words, $a \neq 1$)? I think it only does for $v_p(a) > 0$.
More generally, as soon as the Laurent series is not a power series, and the operator is given by $T \mapsto f(T)$ for some polynomial $f$, the new series can only converge at $T=a$ for $v_p(a) > max \require{enclose} \enclose{horizontalstrike}{\{v_p(c): c\in \mathbb{C}_p , f(c) = 0\}} {\{v_p(c): c\in \mathbb{C}_p \setminus \{0\} , \, f(c) = 0\}} $.
Sketch of proof: Factorise $f(T)= T^{n_0}\prod (T-c_i)^{n_{c_i}}$ over $\mathbb{C}_p$: with partial fractions we can write
$\displaystyle\frac{1}{f(T)} = \frac{p_{0,1}(T)}{T} + ... + \frac{p_{0,n_0}(T)}{T^{n_{0}}} +\sum_i \left(\frac{p_{i,1}(T)}{(T-c_i)} + ... + \frac{p_{i,n_{c_i}}(T)}{(T-c_i)^{n_{c_i}}}\right)$ with polynomials $p_{i,j}$, with some of the polynomials (at least the $p_{i,n_{c_i}}$) $\neq 0$. 
The terms with powers of $T$ in the denominator are not problematic in our context, but for $c \neq 0$, with a (power-of-)geometric series argument,
$\displaystyle\frac{p(T)}{(T-c)^{n}} = \frac{{(-1/c)}^n p(T)}{(1-T/c)^{n}} = {(-1/c)}^n p(T) \left(\sum_{k=0}^\infty \left(\frac{T}{c}\right)^k \right)^n = {(-1/c)}^n p(T) \sum_{k=n-1}^\infty \binom{k}{n-1} \left(\frac{T}{c}\right)^k$
can only converge for $v_p(T) >v_p(c)$. So this is a necessary condition.
