Convergence of the sequence $(a^{1/p^n} + b^{1/p^n})^{p^n}$ in a $p$-adic field Let $K$ be a complete field with respect to a nonarchimedean absolute value such that the residue field has characteristic $p$ and such that the map $x\mapsto x^p$ is surjective on the residue field.
Let $a,b\in K$, and let $a^{1/p^n},b^{1/p^n}$ be compatible sequences of $p$th roots.
How can we see that the sequence $(a^{1/p^n} + b^{1/p^n})^{p^n}$ converges as $n\rightarrow \infty$?
In fact, the limit should be just $a + b$, and philosophically seems similar to the limit
$$\lim_{n\rightarrow\infty}(x^n+y^n)^{1/n} = x+y$$
 A: Without loss of generality, let $a=1$ and $b \in R$ where $R$ is the ring of integers.  Define
$$x_n(b)=(1+b^{1/p^n})^{p^n}. $$
Then, when $n>0$,
$$x_{n+1}(b)-x_n(b)=\left((1+b^{1/p^{n+1}})^{p^n} \right)^p-\left((1+b^{1/p^n})^{p^{n-1}}\right)^p. $$
This factors into
$$\left((1+b^{1/p^{n+1}})^{p^n}-(1+b^{1/p^n})^{p^{n-1}} \right)\left( \sum_{i=0}^{p-1}(1+b^{1/p^{n+1}})^{ip} (1+b^{1/p^n})^{p-1-i} \right). $$
Note that the left term is $x_n(b^{1/p})-x_{n-1}(b^{1/p})$ and so by induction $x_{n+1}(b)-x_n(b)$ has greater valuation  if and only if
$$\sum_{i=0}^{p-1}(1+b^{1/p^{n+1}})^{ip}(1+b^{1/p^n})^{p-1-i} \equiv 0 \pmod{\pi} $$
with $\pi$ the ideal; that is, they become equal when passing to the residue field.  However, in the residue field of characteristic $p$, note that the $p$th power operation (the Frobenius) is an endomorphism, so $(1+b^{1/p^{n+1}})^p \equiv 1+b^{1/p^n}$ and
$$\sum_{i=0}^{p-1}(1+b^{1/p^{n+1}})^{ip}(1+b^{1/p^n})^{p-1-i} \equiv \sum_{i=0}^{p-1}(1+b^{1/p^n})^{p-1} \equiv p(1+b^{1/p^n})^{p-1} \equiv 0. $$
Therefore, we can define
$$x_\infty(b) = \sum_{n=0}^\infty x_{n+1}(b)-x_n(b) $$
as a sum which converges for all sets $\{b^{1/p^n}\}$.  (Note, however, that this will not simply be $1+b$ as you were expecting; in particular, different choices of $b^{1/p}$, etc. will lead to different values of $x_\infty$.)
