Given the sequence $\{ a_n\}$ defined by the (positive and $a_n < 7^n$) solutions of the congruence $x^2 \equiv 2 \mod 7^n$ and $a_{n+1}\equiv a_n \mod 7^n$. e.g. the first one is $a_1 =3$ the second $a_2 = 3 + 1\cdot 7$, $a_3 = 3 + 1 \cdot 7 + 2 \cdot 7^2$, $a_4 = 3 + 1 \cdot 7 + 2 \cdot 7^2 + 6 \cdot 7^3$, etc... I would like to prove that the partial sums $S_n := \sum_{i=1}^n a_i$ will converge to a limit $S$ so that $S^2 = 2$ in the $\mathbb{Q}_7$ norm, by elementary methods. I am interested in a direct solution that does not use advanced methods or results.
So far I observed that the sequence $\{a_n \}$ does converge to $\sqrt{2}$, and this is what most books also prove, this is direct from the fact $(a_n^2-2)\equiv 0 \mod 7^2 \Rightarrow |a_n^2 - 2|_7 = \frac{1}{7^n} \to 0$ as $n \to \infty$. But I cannot see how this will be extended to the partial sums $S_n$.
Alternatively it seems to me that if I can rearrange the infinite sum $S_{\infty}$ into $a_{\infty}$ then the result follows, but I do not know how to perform such rearrangement, that is how to prove that (even at a formal level only), this is stated in a book but is not proven, nor any hint given:
\begin{eqnarray} S_{\infty}&=&3 + (3+ 1 \cdot 7) + (3+ 1 \cdot 7 + 2 \cdot 7^2 ) + (3+ 1 \cdot 7 + 2 \cdot 7^2 + 6 \cdot 7^3 ) + \cdots \\ &=& 3+ 1 \cdot 7 + 2 \cdot 7^2 + 6 \cdot 7^3 + \cdots = a_{\infty} \end{eqnarray}
my try was to reduce by $\mod 7$ (and then try reducing by $\mod 7^m$) a given $S_n^2$, I was aiming to get 2 + something with small norm $|\cdot|_7$ and , but this did not work, I could only prove congruence to $2n^2$ which is neither what I expected nor useful.
\begin{eqnarray} S_n^2 = \sum_{i=1}^{n}a_i^2 + 2 \sum_{i < j}^n a_i a_j \equiv (2n + 2 \sum_{i=1}^{n}\sum_{k=1}^{n-i}a_i a_{i+k} ) \mod 7 \end{eqnarray} then by construction $a_{i+1} \equiv a_i \mod 7$ which implies $a_{i+k} \equiv a_i \mod 7$ hence
$$ S_n^2 \equiv 2n + \sum_{i=1}^{n}\sum_{k=1}^{n-i}a_i^2 \equiv 2n + 2 \sum_{i=1}^{n}\sum_{k=1}^{n-i}2 \equiv 2n^2 \mod 7 $$
I think I am on the wrong track here.
The partial sums do not converge. From $a_1\equiv 3\pmod 7$ we see that $a_n\equiv 3\pmod 7$ for all $n$ and hence $S_n$ is periodically $\equiv 3,6,2,5,1,4,0,3,6,2,5,1,4,0,\ldots\pmod 7$ and cannot converge (whatever $S$ you pick, most partial sums would have $|S_n-S|_7\ge 1$).
Remember that for the convergence of partial sums it is necessary (and in the $p$-adic case also sufficient) that the summdns tend to $0$.