Let $f(x)=\displaystyle{\sum_n a_nx^n}$ be a power series with coefficients in the field of p-adic rationals, $\mathbb{Q}_p$. Let $R_f$ be its radius of convergence and let $f'$ denote its term-by-term derivative.
I have two (related) questions:
$(1)$ Is it the case that $R_f\leq R_{f'}$?
$(2)$ Is it ever possible that the inequality is strict?
We have that $$\frac{1}{R_{f}}=\limsup_{n\to\infty}|a_n|_p^{1/n}$$ and $$\frac{1}{R_{f'}}=\limsup_{n\to\infty}|na_n|_p^{1/n}$$ by the so-called root test.
Now, $$|na_n|_p^{1/n}=|n|_p^{1/n}|a_n|_p^{1/n}$$ I want to somehow show that the $|n|_p^{1/n}$ forces the limit superior to be larger than that of the original function, but am stuck.
As for the second question, my inuition tells me that this should be possible? I'm having trouble ocming up with a solid (counter)example.
Thanks!
Edit: Fixed the multiple typos in the radius of convergence formula.
The root test is the replacement of $\varlimsup$ by the more restrictive $\lim$. Not all sequences have a limit, but they all have a $\varlimsup$ in the extended real numbers.
Example. The sequence $\sqrt[n]{|n|_p}$ has no limit, but it has a $\varlimsup$. Can you compute it?
Remark. The series $f(x) = \sum_{k \geq 0} x^{p^k}$ converges if and only if $|x|_p < 1$, while $f'(x) = \sum_{k \geq 0} p^kx^{p^k-1}$ converges if $|x|_p \leq 1$, so $f'$ converges on a larger set than $f$. This is quite unlike the situation in $\mathbf C$: the complex power series $\sum_{n \geq 1} x^n/n^2$ converges if and only if $|x| \leq 1$, but the series $\sum_{n \geq 1} x^{n-1}/n$ does not converge at $x = 1$: the derivative converges on a smaller set than the original series.