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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)

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  • $\begingroup$ By the way: I would be happy to know of a more elementary proof, if there is one. $\endgroup$ – A.P. Mar 24 '15 at 22:25
  • $\begingroup$ The claim is too strong. For example, if $\alpha_n = -1$, then the product need not converge. $\endgroup$ – Sangchul Lee Mar 24 '15 at 23:00
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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 $|\alpha_n|_p=1$.

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  • $\begingroup$ I thought about this, but I didn't know how to prove that it it isn't ultimately periodic. How do you do it? Also, if this is indeed the case, can you tell me where is the flaw in my argument? $\endgroup$ – A.P. Mar 24 '15 at 22:51
  • $\begingroup$ @A.P. Consider the sum until $n=2^k$, you will find no period. Now until $2^{k+1}$, and induct.I have not worked out the details, but it should work. For your second question, I cannot really answer it because I lack the knowledge ;) $\endgroup$ – chubakueno Mar 24 '15 at 23:17
  • $\begingroup$ Aren't the sums in $(*)$ and in the next formula equivalent? Or, in other words, why do you justify the equivalence with "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"? $\endgroup$ – A.P. Mar 25 '15 at 12:34
  • $\begingroup$ This is to justify that, in the case where $p^m$ divides $\ell$, $\ell_m$ is being changed from $p^m$ to $0$. Inside the braces, $2^{\ell_m}/(2^{p^m}-1)$ will then change from $2^{p^m}/(2^{p^m}-1)$ to $1/(2^{p^m}-1)$. This does not change the overall value of the expression however as these two quantities differ by $1$ and we are taking the fractional part. $\endgroup$ – Polichinelle Mar 25 '15 at 20:48
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Your claim is too strong to be true, but the particular product you are interested may be analyzed in a slightly different way. Indeed, let $\alpha_n$ be a zero of $x^{2^{n-1}} - n$ in $\Bbb{C}_p$. Then clearly we should have

$$ \prod_{n=1}^{\infty} \alpha_n \quad \text{coverges in } \Bbb{C}_p \quad \Longrightarrow \quad \lim_{n\to\infty} \alpha_n = 1 \quad \text{in } \Bbb{C}_p. $$

Now, since $|\alpha_n|_p \neq 1$ if $p \mid n$, for such $n$ we must have

$$ |\alpha_n - 1|_p = \max\{|\alpha_n|_p, |1|_p\} = 1 $$

from the Krull sharpening, and hence the product does not converge.

This does not tell us whether the product is rational or not, but the following simplification might have a chance to be useful.

$$ \sum_{n=1}^{\infty} v_p(n)x^n = \sum_{n=1}^{\infty} \frac{x^{p^n}}{1 - x^{p^n}}. $$

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  • $\begingroup$ Thanks, you made clear why my argument fails. I would accept both answers if I could... $\endgroup$ – A.P. Mar 25 '15 at 12:36

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