In Silverman's Arithmetic of Elliptic Curves, Lemma IV.6.3(a) states that
Let $R$ be a ring of characteristic $0$, complete with respect to a discrete valuation $v$, and let $p\in\mathbb{Z}$ be a prime with $v(p)>0.$ (a) Let $f$ be a power series of the form $f(T) = \sum_{n=1}^\infty \frac{a_n}{n} T^n$ with $a_n\in R$. If $x\in R$ satisfies $v(x)>0$, then the series $f(x)$ converges in $R$.
The proof is given as follows:
${\it{Proof.}}$ For a general term of $f(x)$ we have $$ v(a_nx^n/n) \geq v(a_n) + nv(x) - v(n) \geq nv(x) - v(n) \geq nv(x) - (\log_p n)v(p). $$ This last expression goes to $\infty$ as $n\to \infty$. Since $v$ is nonarchimedean and $R$ is complete, the series $f(x)$ converges.
My question about this proof is that nowhere does it show that $v(a_nx^n/n)\geq 0$ for all $n$ -- I think this is needed for us to know that $f(x)$ converges in $R$ (instead of just converging in the fraction field of $R$.) Later (Theorem IV.6.4(b), where $f$ is the formal logarithm), under the assumption that $v$ is normalized (that is, $v(\mathrm{Frac}(R)^\times) = \mathbb{Z}$), Silverman thinks that it is obvious that if $v(x)\geq r$, then $v(f(x))\geq r$. But I do not find it obvious. Could anyone please explain why the convergence is in $R$? Thanks in advance!
Thoughts. Using the lower bound obtained in the proof given above, to show that $v(a_nx^n/n)>0$ it suffices to show that $\displaystyle v(x) > \frac{\log_p n}{n} v(p)$ for all $n\geq 1$. We clearly have that $p=\mathrm{char}(R/\mathfrak{m})$, where $\mathfrak{m}=\{x\in R| v(x)>0\}$. If one can show that $\color{red}{\textrm{$v(p)=\min\{v(x)|x\in \mathfrak{m}\}$}}$, then we are done since $\displaystyle \max_n \frac{\log_p n}{n} \leq \frac{1}{e \log p} \leq \frac{1}{e \log 2} < 1$ and thus $\displaystyle v(x) \color{red}{\geq} v(p) > \frac{\log_p n}{n} v(p).$ As for Theorem IV.6.4(b), if $v$ is normalized and my thoughts above are correct, then $\color{red}{v(p)=1}$ and $$ v(a_n x^n / n) \geq nv(x)-(\log_p n) v(p)) \geq v(x) - \frac{\log_p n}{n} v(p) > r-1,$$ from which it will follow that $v(a_n x^n / n) \geq r$ for all $n$.
Update: Thanks to @Torsten Schoeneberg and his comment, this is a slight mistake of the book.