p-adic field or Non-Archimedian valued field $ \mathbb{Q}_p$:
If a power series of the form $ \ \sum_{n \geq 0} a_n (x-3)^n $ in $ \mathbb{Q}_p$ has radius of convergence $ \ R=p^{-\frac{2}{p-1}} $ . , then $$ |x-3|_p<p^{-\frac{2}{p-1}} \ , \cdots \cdots (1)$$ where $ p$ is prime.
How to write disc of convergence in $ (1) $ in the following form $$ |x|_p < K $$ where $K$ is to be determined.
Answer:
$|x|_p=|(x-3)+3|_p \leq |x-3|_p+|3|_3 <p^{-\frac{2}{p-1}}+|3|_3 =1+p^{-\frac{2}{p-1}}=K, say$
I need confirmation of my work.
Help me
As mentioned in the comment, this is impossible. If there were a positive $K$ with $\{x: |x|_p < K\} = \{x: |x-3|_p<p^{-\frac{2}{p-1}} \}$ , then necessarily $x=0$ would have to be in the RHS set, in other words we would have $|-3|_p<p^{-\frac{2}{p-1}}$. But we have
$|-3|_p = |3|_p = \begin{cases} 1 \not <p^{-\frac{2}{p-1}} \qquad\text{ if }\; p \neq 3\\ 3^{-1} \not <3^{-\frac{2}{3-1}} \qquad \text{ if } p=3.\end{cases}$
In other words, the series does not converge at $0$, hence in no disk around (i.e., in particular containing) $0$.
Note that if $R$ were strictly greater than $1$ resp. $1/3$, you could just choose $K=R$, because then $ 0 \in \{x: |x-3|_p<R \}$, and by the strong triangle inequality one has
$$|x_1-x_2|_p < R \Rightarrow \qquad [\text{for all } y, |y-x_1|_p < R \Leftrightarrow |y-x_2|_p >R]$$
or in other words, any point contained in an open ball of a certain radius can serve as centre of that ball (here we would apply that to $x_1 =0$ and $x_2 = 3$).