Convergence of P-adic series How can I prove that $\sum_{i = 0}^\infty 2^i$ doesn't converge with respect to $|\cdot|_3$ without resorting to the fact that $\sum a_i$ converges wrt $|\cdot|_p$ if and only if $|a_i|_p \to 0$? Is it even possible?
 A: The only thing that $\sum_0^\infty 2^n$ can converge to is $-1$. But
this series does not converge $3$-adically to $-1$. If it did then
$$\left|-1-\sum_{n=0}^N 2^n\right|_3<1$$
for all large enough $N$. This means that
$$\sum_{n=0}^N 2^n\equiv-1\pmod 3\tag{*}$$
for all large enough $N$. But $3\nmid 2^n$ for all $n$, so if (*) holds
for $N$, it fails for $N+1$. Contradiction.
A: We are interested in the convergence of the sequence $\{S_n\}$, where $S_n=\sum_{i=0}^n2^i$.
Notice that $|S_n-S_{n-1}|_3=|2^n|_3=1$, valid for all $n$.
Now go for a contradiction and suppose that $L$ is a limit, i.e. $L=\lim_nS_n$. This means that given $\varepsilon>0$, there is $N$ such that for all $n\ge N$, the inequality $|L-S_n|_3<\varepsilon$ holds. In particular, taking $\varepsilon=\frac12$,  there’s $N$ such that for all $n\ge N$, $|L-S_n|<\frac12$.
Now look at $|S_N-S_{N+1}|_3$. We have:
\begin{align}
|S_N-S_{N+1}|_3&=|L-S_{N+1}+(S_N-L)|_3\\
&\le\max\bigl(|L-S_{N+1}|_3,|L-S_n|_3\bigr)\\
&<\frac12\,,
\end{align}
giving us the required contradiction.
