p-adic binomial series I want to show that the sequence $x_n=1+\sum_{i=1}^{n}\begin{pmatrix}1/2\\i\end{pmatrix}15^i$ converges to $4$ w.r.t $|\cdot|_3$ and $-4$ w.r.t $|\cdot|_5$. Showing the sequence converges in $\mathbb{Q}_p$ is easy as it suffices to show that the sequence is Cauchy. On the other hand, I don't know how to show that the sequences to $4$ and $-4$ w.r.t the two absolute values.
(Example with the $|\cdot|_3$
I want to show that $|x_n-4|_3=\frac{1}{3^{n+1}}$.
My idea was to use the induction steps,
1.$(x_n-4)/3^{n+1}\in 2+3\mathbb{Z}_3$
2.$(x_{n+1}-x_n)/3^{n+1}\in 1+3\mathbb{Z}_3$
To then induct and say $3^{n+2}|x_{n+1}-4$
The second statement can be proved directly without much effort by noting that $\displaystyle \frac{x_{n+1}-x_n}{3^{n+1}}=\begin{pmatrix}1/2\\n+1\end{pmatrix}5^{n+1}=\frac{(1/2)\times(-1/2)\times(-3/2)...((-1-2n)/2)}{1\times 2\times 3...(n+1)}\times 5^{n+1}=\frac{1\times(-1)\times(-3)\times...(1-2n)}{1\times2\times 3\times...(1+n)}\times(5/2)^{n+1}\equiv\frac{1\times 2\times 3\times...(n+1)}{1\times 2\times 3\times ...(n+1)}\times (2/2)^{n+1}\equiv 1(\textrm{mod }3)$.
However, the first statement is much more difficult to prove and I have no idea how to show the convergence.
So my question is I am taking the correct approach? How does one show that a particular p-adic series converges to a particular value in general as I feel like I ought to invoke some number theoretic property about divisibility in powers of $p^n$. 
 A: I think the most useful way of looking at this situation is that the series
$$
H(x)=1+\sum_1^\infty\binom{1/2}nx^n
$$
is the Binomial Series for $(1+x)^{1/2}$, as @Hurkyl has already noted. Please note that the only denominators in the coefficients are powers of $2$: they are $p$-integral for all $p\ne2$. This means that the series is $p$-adically convergent for all values of $x$ with $\vert x\vert_p<1$.
I’ll omit the argument that the series is a formal square root of the series $(1+x)$. That is, that if we call the ring $R=\Bbb Z[1/2]$, we have a formal identity $H(x)^2=1+x$, in the formal power-series ring $R[[x]]$.
But granting that, you have to notice that when you evaluate $x$ to an element of $z\in p\Bbb Z_p$, you certainly get a result that is $\equiv1\pmod p$. In particular, in either $\Bbb Z_3$ or $\Bbb Z_5$, you get the result that $H(z)$ not only is a square root of $1+z$, but it’s also congruent to 1 modulo p.
That’s what does it: the square root of $16$ that’s $\equiv1\pmod3$ is $+4$, while the square root of $16$ that’s $\equiv1\pmod5$ is $-4$.
A: 
Theorem (Legendre's formula): For any prime $p$ and natural number $n$, the $p$-adic valuation of $n!$ is given by
  $$ v_p(n!) = \sum_{i=1}^{\infty}  \left\lfloor \frac{n}{p^i} \right\rfloor < \frac{n}{p-1}$$
Corollary: For any $x \in \mathbf{C}_p$, if $|x|_p < p^{-1/(p-1)}$, then
  $$\lim_{n \to \infty} \frac{x^n}{n!} = 0 $$
In particular, if $p > 2$, then $|p|_p = p^{-1} < p^{-1/(p-1)}$.

An infinite sum over the $p$-adics converges if and only if its terms converge $p$-adically to zero (prove this!), so this can be used to show convergence.

To actually find the value of the sum, it turns out that the series you are considering has a closed form:
$$ \sqrt{1 + x} = 1 + \sum_{n=1}^{\infty} \binom{1/2}{n} x^n $$
Given this, all you need to do is to determine how many digits of precision you need to distinguish between $4$ and $-4$, and then determine at which point in the summation all of the remaining terms are insignificant.
