p-adic binomial series

Question

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 $p^n$.

Answer 1

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 $-4$.

Answer 2

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.

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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.