Fix $p \in \Bbb{Z}$ a prime number and let $v_p$ be the usual $p$-adic valuation on $\Bbb{Q}$. I would like to know if $$ \sum_{n = 1}^{\infty} \frac{v_p(n)}{2^{n-1}} $$ is a rational number.
I think it should be, based on the following assumption (it seems reasonable to me, but I couldn't find a reference):
Claim: If $(\alpha_n)$ is a sequence of algebraic integers, then $$ \prod_{n = 1}^{\infty} \alpha_n $$ converges in $\Bbb{C}_p$ with respect to $\left|\cdot\right|_p$ if and only if $\left| \alpha_n \right|_p \to 1$ for $n \to \infty$. (Here $\left|\cdot\right|_p$ is the unique normalised extension to $\Bbb{C}_p$ of the usual $p$-adic absolute value on $\Bbb{Q}$.)
Then consider the sequence of algebraic integers $(\alpha_n) = (n^{1/2^{n-1}})$. Since $$ \left| \alpha_n \right|_p = p^{-v_p(n)/2^{n-1}} \to 1 $$ then by the claim (if it is true) it follows that $$ \prod_{n = 1}^{\infty} n^{1/2^{n-1}} $$ converges to some $\alpha \in \Bbb{C}_p$. Finally, by the definition of the extension of $\left|\cdot\right|_p$ from the algebraic closure of $\Bbb{Q}_p$ to $\Bbb{C}_p$, we know that $\left| \alpha \right|_p$ is the limit of the absolute values of the partial products, i.e. $$ \left| \alpha \right|_p = \lim_{k \to \infty} \left| \prod_{n = 1}^{k} n^{1/2^{n-1}} \right|_p = p^{-\sum_{n = 1}^{\infty} v_p(n) 2^{-n+1}} $$ and by Proposition 1.3 and Proposition 2.1.1 of [1], chapter 3, we know that $\left|\cdot\right|_p : \Bbb{C}_p \to p^{\Bbb{Q}}$, so $$ -\sum_{n = 1}^{\infty} \frac{v_p(n)}{2^{n-1}} $$
Note: I'm asking because I come from algebraic number theory and this is the first time I have to do with $p$-adic analysis, plus I have never been much good at classical analysis: so please understand any blunder I may have made in this regard (and point them out, with a thorough explanation if possible or with some references)!
[1] Alain Robert, A Course in $p$-adic analysis (Google Books)
The binary expansion of $r_p:=\sum_{n\ge 1} v_p(n) 2^{-(n-1)}$ does not repeat. Therefore, this number is irrational.
To prove this, let $\{x\}:=x-\lfloor x\rfloor$ be the fractional part of $x$. If the binary expansion of $r_p$ repeated, then the sequence $(\{2^k r_p\})_{k\ge 0}$ would also eventually be periodic. Now \begin{eqnarray*} \{2^k r_p\}&=&\{\sum_{n\ge 1} \frac{v_p(n)}{2^{n-1-k}}\}\\ &=& \{\sum_{n\ge k+1} \frac{v_p(n)}{2^{n-1-k}}\}\\ &=& \{\sum_{n\ge 0} \frac{v_p(n+k+1)}{2^{n}}\}, \end{eqnarray*} so in this case the sequence $r'_{p,0}$, $r'_{p,1}$, $\dots$, where $$r'_{p,\ell}:= \{\sum_{n\ge 0} \frac{v_p(n+\ell)}{2^n}\},$$ will eventually be periodic. Therefore, it can only assume finitely many values.
Now, \begin{eqnarray*} r'_{p,\ell} &=& \{\sum_{m\ge 1} \sum_{n\ge 0:\ \ p^m {\rm \ divides \ } n+\ell} 2^{-n}\}. \ \ \ (*) \end{eqnarray*} If $\ell_m:=p^m$ when $p^m$ divides $\ell$, and otherwise $\ell_m$ is the remainder of $\ell$ when divided by $p^m$, then we can rewrite $(*)$ as $$ r'_{p,\ell}=\{\sum_{m\ge 1} \frac{2^{\ell_m}}{2^{p^m}-1}\}. $$ In fact, since we are taking the fractional part, we can always take $\ell_m$ to be the remainder of $\ell$ when divided by $p^m$ in the above. Therefore, if $\ell=p+p^3+\cdots+p^{2r+1}$, then $r'_{p,\ell}$ is equal to $$ \{\frac{1}{2^p-1} +\sum_{1\le m\le r+1} \frac{2^{p+p^3+\cdots+p^{2m-1}}}{2^{p^{2m}}-1} +\frac{2^{p+p^3+\cdots+p^{2m-1}}}{2^{p^{2m+1}}-1} +\sum_{m> 2r+3} \frac{2^{p+p^3+\cdots+p^{2r+1}}}{2^{p^m}-1}\}.\ \ (**) $$ The quantity inside the braces in $(**)$ is strictly increasing with $r$.
It is bounded above by \begin{eqnarray*} \frac{1}{2^p-1} +\sum_{m\ge 1} \frac{2^{p+p^3+\cdots+p^{2m-1}}}{2^{p^{2m}}-1} +\frac{2^{p+p^3+\cdots+p^{2m-1}}}{2^{p^{2m+1}}-1}. \end{eqnarray*} This is a rapidly convergent series whose sum can easily be estimated to be less than $1$. This means that $(**)$ assumes infinitely many values, contradicting the periodicity of $r'_{p,\ell}$.
The decimal expansion of $r_2/2+1$ is OEIS A048649. Also, there is another irrationality proof for this number in: S. W. Golomb, "On the sum of the reciprocals of the Fermat numbers and related irrationalities", Canad. J. Math., 15 (1963), 475-478..
The flaw in your argument is that your Claim is too strong. As pointed out in the other answer, to have $\prod_{n\ge 1} \alpha_n$ converge, you need to have $\alpha_n-1\to 0$, and therefore, from some point on, $|\alpha_n-1|_p<1$, which means that $|\alpha_n|_p=1$.