---
abstract: 'We show that the Lambert series $f(x)=\sum d(n) x^n$ is irrational at $x=1/b$ for negative integers $b < -1$ using an elementary proof that finishes an incomplete proof of Erdős.'
author:
- 'J. Vandehey'
title: On an incomplete argument of Erdős on the irrationality of Lambert series
---

Introduction
============

Chowla [@chowla] conjectured that the functions $$f(x) = \sum_{n=1}^\infty \frac{x^n}{1-x^n} \qquad \text{and} \qquad g(x) = \sum_{n=1}^\infty \frac{x^n}{1-x^n}(-1)^{n+1}$$ are irrational at all rational values of $x$ satisfying $|x|<1$. For such $x$ the above functions may be rewritten as $$f(x) = \sum_{n=1}^\infty d(n) x^n \qquad \text{and} \qquad g(x) = \frac{1}{4}\sum_{n=1}^\infty r(n)x^n,$$ where $d(n)$ is the number of divisors of $n$ and $r(n)$ is the number of representations of $n$ as a sum of two squares.

Erdős [@erdos1] proved that for any integer $b>1$, the value $f(1/b)$ is irrational. He did so by showing that $f(1/b)$ written in base $b$ contains arbitrary long strings of $0$’s without terminating on $0$’s completely. If we take $b<-1$ to be a negative integer, then Erdős’ methods show by the same method that $f(1/b)$ in base $|b|$ contains arbitrary long strings of $0$’s; however, Erdős claims without proof that showing it will not terminate on $0$’s can be done using similar methods. It is not clear what method Erdős intended, and in later papers (including his review of similar irrationality results [@erdos2]) Erdős only refers to proving the case of positive $b$.

Since then, several proofs have been offered for the irrationality of the $b<-1$ case and far more general theorems besides. Much credit is often given to Bezivin [@bezivin] and Borwein [@borwein] for proving the first major generalizations of these results; and other results can be often be found in the literature under the term of the $q$-analogue of the logarithm or, simply, the $q$-logarithm. However, these results are proved using entirely different techniques than what Erdős uses and leaves open the question of whether his method could have finished the proof.

Erdős’ method can be extended to the following stronger result with a virtually identical proof.

\[thm:erdos\] Let $b>1$ be a positive integer and $\mathcal{A}$ be any finite set of non-negative integers. Then for any sequence $\{a_n\}_{n=1}^\infty$ taking values in $\mathcal{A}$ such that the sequence does not end on repeated $0$’s, we have that $$\sum_{n=1}^\infty d(n) \frac{a_n }{b^n}$$ is irrational.

Theorem \[thm:erdos\] has the following curious corollary. Let $a_n(x)$ be the $n$th base $b$ digit of a number $x$ in $(0,1)$. (If $x$ has two base $b$ expansions, then we chose the one which does not end on repeated $0$’s.) Then the map $$x=\sum_{n=1}^\infty \frac{a_n(x)}{b^n} \longmapsto \sum_{n=1}^\infty  d(n) \frac{a_n(x)}{b^n}$$ has its image in $\mathbb{R}\setminus \mathbb{Q}$ and is also continuous at all $x$ that do not have a representation as a finite base $b$ expansion.

We could replace the condition that $a_n$ be in the finite set $\mathcal{A}$ with a restriction that $0\le a_n \le \phi(n)$ for some sufficiently slowly growing integer-valued function $\phi$. It would be interesting to know what the fastest growing $\phi$ for which the Theorem \[thm:erdos\] holds would be.

In this paper, we will prove the following extension of Theorem \[thm:erdos\].

\[thm:main\] Let $b>1$ be a positive integer and $\mathcal{A}$ be any finite set of integers *that does not contain* $0$. Then for any sequence of $\{a_n\}_{n=1}^\infty$ taking values in $\mathcal{A}$, we have that $$\sum_{n=1}^\infty d(n)\frac{a_n}{b^n}$$ is irrational.

The new ingredient to extend Erdős’ method is finding arbitrarily long strings of zeros *that are known to be preceded by a non-zero number*, and to find these strings arbitrarily far into the base $|b|$ expansion.

In particular, by taking $a_n=(-1)^n$, this proves that $f(1/b)$ is irrational for negative integers $b<1$ as well, completing Erdős’ proof.

Proof of Theorem \[thm:main\]
=============================

We will require a result mentioned by Alford, Granville, and Pomerance [@agp p. 705]. The function $\pi(N;d,a)$ equals the number of primes up to $N$ that are congruent to $a$ modulo $d$.

\[prop:agb\] Let $0<\delta<5/12$. Then there exist positive integers $N_0$ and $\overline{\mathcal{D}}$ dependent only on $\delta$, such that the bound $$\pi(N;d,a) \ge \frac{N}{2\varphi(d) \log N}$$ holds for all $N>N_0$; all moduli $d$ with $1 \le d \le N^\delta$, except, possibly for those $d$ that are multiples of some element in $\mathcal{D}(N)$, a set of at most $\overline{\mathcal{D}}$ different integers that all exceed $\log N$; and all a relatively prime to $d$.

We begin our proof much as Erdős did his. Let $b \ge 2$ be a fixed positive integer, let $\mathcal{A}$ be a finite set of integers that does not contain $0$, and let $N$ be a large positive integer that is allowed to vary. Define $k$ in terms of $N$ by $$k=k(N):=\lfloor \left( \log{N} \right)^{1/10} \rfloor.$$ Let $j_0$ be a fixed integer, independent of $N$, so that $2 \max_{a\in \mathcal{A}} |a| / b^{j_0} <1$.

Let $0<\delta<5/12$ be some sufficiently small fixed constant, and let $N_0=N_0(\delta)$ and $\overline{\mathcal{D}}=\overline{\mathcal{D}}(\delta)$ be the corresponding constants from Proposition \[prop:agb\]. Let $N_1 >N_0$ be large enough so that for any $N> N_1$, the interval $((\log N)^2, 2 (\log N)^2)$ cotains at least $u+\overline{\mathcal{D}}$ primes, where $u=u(N)=k(k-1)/2$. In addition, for such $N>N_1$, let $\mathcal{D}(N)$ be the set of exceptional moduli from Proposition \[prop:agb\]. Since we assume that $\delta$ is constant, $|\mathcal{D}(N)|\le \overline{\mathcal{D}}$ is bounded.

For each $D$ in $\mathcal{D}(N)$, let $\tilde{p}_D$ denote the smallest prime strictly greater than $(\log N)^2$ that divides $D$, if such a prime exists, and then let $p_1< p_2< \dots <p_u$ be the smallest $u$ primes strictly greater than $(\log N)^2$ that are not equal to $\tilde{p}_D$ for any $D\in \mathcal{D}(N)$; by assumption on $N$, we have that each such $p_i$ is less than $2(\log N)^2$. Finally, let $$A:= \prod_{i=1}^{ j_0(j_0-1)/2}p_i^b \prod_{i=j_0(j_0+1)/2+1 }^{ u} p_i^b,$$ so that, in particular, $A$ is not a multiple of any $D$ in $\mathcal{D}(N)$; moreover, provided $N$ is sufficiently large, we have $$A<(2(\log N)^2)^{bk(k-1)/2} \le N^\delta.$$

If we set $$A= \prod_{1 \le i \le k^2} p_i^b$$ then since $p_i < 2 \left( \log{N} \right)^2$ for $i \le k^2$ and sufficiently large $N$, we have that—again for sufficiently large $N$— $$A<e^{(\log{N})^{1/4}}.$$

By the Chinese remainder theorem, there exists an integer $r$, with $0 \le r \le A-1$, such that $$r+j\equiv \prod_{i=j(j-1)/2+1}^{j(j+1)/2} p_i^{b-1} \pmod{\prod_{i=j(j-1)/2+1}^{j(j+1)/2} p_i^{b}}, \qquad 0 \le j \le k-1, j\neq j_0.$$ (The exception $j\neq j_0$ marks the key difference between this proof and Erdős’.) Since all the $p_i$’s are bounded below by $(\log N)^2$, we have that $r$ necessarily tends to infinity as $N$ does, although possibly much slower.

With this value of $r$, for any integer of the form $$r+mA, \qquad 0 \le m <\lfloor N/A \rfloor,$$ we have that $$\label{eq:congruence}
d(r+mA+j) \equiv 0 \pmod{b^{j+1}}, \qquad 0 \le j <k,\ j \neq j_0$$ by the multiplicity of $d(\cdot)$. Moreover, $r+j_0$ is relatively prime to $A$, since each $p$ dividing $A$ also divides some $r+j$, with $0 \le j <k,$ $j \neq j_0$; the largest $j$ can be is $k\le (\log N)^{1/10}$, but all primes dividing $A$ are at least $(\log N)^2$.

We can also apply Proposition \[prop:agb\] to see that $$\label{eq:agbcor}
\pi (N,A,r+j_0)\ge \frac{N}{2\varphi(A)\log N}.$$

Erdős in [@edros1] also proved the following result, which we give here without reproof. (While our construction of $A$ and $r$ are different from Erdős’, they satisfy all the requirements for Erdős’ proof technique to still hold.)

\[lem:erdos\] With $A$, $r$, $b$, and $k$ all as above, the number of $m < \lfloor N/A \rfloor$ such that $$\sum_{n> r+k+mA} d(n)\frac{1}{b^n} > \frac{1}{b^{r+k/2+mA}}$$ is less than $$\frac{10cN(\log{N})^2}{A2^{k/4}}$$ for some constant $c$ independent of all variables.

Regardless of how large $c$ is, we have, for sufficently large $N$, that $$\frac{N}{2\varphi(A)\log N} \ge \frac{10cN(\log{N})^2}{A2^{k/4}}.$$ Therefore, by combining Lemma \[lem:erdos\] with and , we see that for sufficiently large $N$ there exists some $m_0 < \lfloor N/A\rfloor$ such that $$\label{eq:upper}
b^{j+1} | d(r+m_0 A+j) , \qquad 0 \le j <k, j \neq j_0,$$ $$\label{eq:middle}
r+m_0 A + j_0 \text{ is prime},$$ and $$\label{eq:lower}
\sum_{n> r+k+m_0 A} \left| d(n) \frac{a_n }{b^n} \right| \le \frac{\max_{a\in \mathcal{A}} |a|}{b^{r+k/2+mA}}.$$

Now consider a particular sequence $(a_n)$ with each $a_n \in \mathcal{A}$ together with the sum $$\sum_{n=1}^\infty d(n) \frac{a_n }{b^n} .$$ By , the partial sum $$\sum_{\substack{n \le r+k+m_0 A \\  n \neq r+j_0+m_0 A}} d(n) \frac{a_n }{b^n},$$ when written in base $b$, has its last non-zero digit in the $(r-1+m_0A)$th place or earlier.[^1] In addition, by , the partial sum $$\sum_{n > r+k+m_0 A} d(n) \frac{a_n }{b^n}$$ when written in base $b$ has its first non-zero digit in the $$(r+k/2+m_0 A-\lceil \log_b \max_{a\in \mathcal{A}} |a| \rceil)\text{th}$$ place or later. The number $$d(r+j_0+m_0A) \frac{a_{r+j_0+m_0A} }{b^{r+j_0+m_0A}}=\frac{2a_{r+j_0+m_0A} }{b^{r+j_0+m_0A}}$$ when written in base $b$ has its non-zero digits only between the $(r+m_0  A)$th and $(r+j_0+m_0 A)$th place, and it has at least one such non-zero digit. Thus the full sum has a string of at least $k/2+O(1)$ zeroes immediately preceded by a non-zero digit starting somewhere between the $(r+m_0 A)$th and $(r+j_0+m_0 A)$th place.

So as $N$ increases to infinity, we can find arbitrarily long strings of $0$’s (which corresponds to $k$ increasing to infinity) immediately preceded by a non-zero digit, and we find these strings arbitrarily far out in the expansion (since $r$ also tends to infinity). The base $b$ digits cannot therefore be periodic and hence the sum is irrational. This completes the proof.

$$\begin{aligned}
n &\equiv q_1^{b-1} \pmod{q_1^b}\notag \\
n +2 &\equiv (q_2 q_3 q_4)^{b-1} \pmod{(q_2 q_3 q_4)^b} \notag \\
n +3 &\equiv (q_5 q_6 q_7 q_8)^{b-1} \pmod{(q_5 q_6 q_7 q_8)^b} \label{eq:congruence} \\
&\cdots \notag \\
n+k-1 &\equiv (q_u q_{u+1} \ldots q_{u+k-1})^{b-1} \pmod{(q_u q_{u+1} \ldots q_{u+k-1})^b}. \notag\end{aligned}$$ The absence of a congruence for $n+1$ is intentional and marks a key difference between the proof here and Erdős’ proof.

Let $D=D(N)$ be the exceptional modulus from Proposition \[prop:agb\]. If $D$ and $A$ share a prime factor $p_j$, let $B=A/p_j^t$; otherwise, let $B=A/p_1^t$. Then, we have that $D$ does not divide $B$.

Let $q_1, q_2, \ldots$ be the various distinct prime factors of $B$ in no particular order. Then, by the Chinese remainder theorem, there exists an $x$ modulo $B$ which satisfies the following congruences: $$\begin{aligned}
x &\equiv q_1^{t-1} \pmod{q_1^t}\\
x +2 &\equiv (q_2 q_3 q_4)^{t-1} \pmod{(q_2 q_3 q_4)^t}\\
x +3 &\equiv (q_5 q_6 q_7 q_8)^{t-1} \pmod{(q_5 q_6 q_7 q_8)^t}\\
&\cdots\\
x+k-1 &\equiv (q_u q_{u+1} \ldots q_{u+k-1})^{t-1} \pmod{(q_u q_{u+1} \ldots q_{u+k-1})^t}\end{aligned}$$ where $u = k(k-1)/2 -1$. The absence of a congrence for $x+1$ is intentional.

With this choice of $x$ modulo $B$, we have that any integer of the form $$x+yB, \qquad 0 \le y < \lfloor N/B \rfloor$$ we have by the multiplicity of $d(n)$ that $$d(x+j+yA) \equiv 0 \pmod{t^{j+1}}$$ for all $0 \le j < k$, $j \neq 1$.

We have that $x+1$ is relatively prime to $B$, since each $q_i$ divides a number of the form $x+j$ with $0\le j <k$, $j\neq 1$, and since each $q_i$ is larger than $k$. By construction, $D$ does not divide $B$ so we can apply Proposition \[prop:agb\] to see that $$\pi(N,B,x+1) \gg \frac{N}{\phi(B) \log{N}}$$

We furthermore can apply the above theorem to see how often $x+1+yB$ is prime, since $(x+1,B)=1$ (there exists $0 \le j < k$, $j \neq 1$ for which a given prime dividing $B$ also divides $x+j$ and since $k< \log{n}$ and each prime dividing $B$ is greater than $\log{n}^2$ the result follows) and $D_0 \nmid B$. Then, since $B < n^{\delta}$ for any $\delta$ provided $n$ is sufficiently large, we then have that $$\pi (n,B,x+1) > \frac{n}{2 \phi(B) \log{n}}.$$

Combining this with the above work, we can see that for any large enough $n$ there exists some $y_0>1$ such that $$\sum_{r> x+k+y_0 B} \left| \frac{d(r)}{t^r} \right| < \frac{1}{t^{x+k/2+y_0 B}}$$, $x+1+y_0 B$ is prime, and for all $0 \le j < k$, $j \neq 1$, we have that $t^{j+1} | d(x+j+y_0 B)$.

Erdős himself proved that for any large enough $n$ there exists some $B$ and $y_0>1$ such that $$\sum_{r> x+k+y_0 B} \left| \frac{d(r)}{t^r} \right| < \frac{1}{t^{x+k/2+y_0 B}}$$, and for all $0 \le j < k$, we have that $t^{j+1} | d(x+j+y_0 B)$. To distinguish these two, we call the previous statement fact 1, and this statement fact 2.

This allows us to prove the following:

For any integer $t>1$, the function

$$f(x)=\sum_{r=1}^{\infty} \frac{x^r}{1-x^r} = \sum_{r=1}^{\infty} d(r) x^r$$

is irrational for $x=-1/t$.

Consider the number $a=\sum_{r=1}^{\infty} d(r) (-1/t)^r$. If $a$ is rational, then the fractional parts of $ta,t^2 a, t^3a,\ldots$ should eventually be zero or cyclic depending on whether or not the denominator of $a$ is a power of $t$ or not.

So given a large $n$, consider fact 2 above and multiply $a$ by $t^{x+y_0 B-1}$. Clearly the first $x+y_0 B-1$ terms become integers so are not considered in the fractional part. And the terms $r=x+j+y_0 B$ for $0 \le j < k$ will also be integers since the power of $t$ that divides the numerator will cancel the remaining part of the denominator after multiplying by $t^{x+y_0 B-1}$. Thus the fractional part of $t^{x+y_0 B-1} a$ equals $$t^{x+y_0 B-1} \sum_{r> x+k+y_0 B} \frac{d(r)}{(-t)^r} < t^{-1-k/2},$$ which can be made arbitrarily small by letting $n$ be large. Thus we must fall into the case where $ta,t^2 a, t^3a,\ldots$ is eventually zero.

However, by fact 1, we can find another even larger $n'$ so that its corresponding $x'+y'_0 B'$ is larger than the $n$ from the previous paragraph. So when we multiply $a$ by $t^{x'+y'_0 B' - 1}$, the fractional part equals $$t^{x+y_0 B-1} \left( \frac{d(x'+y'_0 B'+1)}{(-t)^{x'+y'_0 B'+1}} + \sum_{r> x+k+y_0 B} \frac{d(r)}{(-t)^r} \right) \pmod{1}$$ but this first term is just $\pm 2/t^2$ while the latter sum tends to zero. Thus $ta,t^2 a, t^3a,\ldots$ is not eventually zero, so $a$ cannot be rational.

Now consider an arbitrary number in $(0,1]$ and its base $t$ expansion $\sum_{r=1}^{\infty} a_r / t^r$, $a_r \in {0,1,\ldots,b-1}$. If a number has two different expansions (ending on infinite $0$’s or $(b-1)$’s) we will chose the infinite $(b-1)$’s as the proper expansion, so all numbers have infinite sums.

So now we ask, when will $\sum_{r=1}^{\infty} d(r) a_r / t^r$ be irrational?

The answer here is always. We simply apply fact 2 again with arbitrary large $n$ to see that $\left\{ t^{x+y_0 B-1} \sum_{r=1}^{\infty} d(r) a_r / t^r \right\} = t^{x+y_0 B-1} \sum_{r> x+k+y_0 B} d(r)a_r/ t^r$ and this last value is strictly between $0$ and $(b-1)/t^{x+k/2+y_0 B}$. Thus the $t$-power multiples of the sum $\sum_{r=1}^{\infty} d(r) a_r / t^r$ have fractional parts that can be arbitrarily close to $0$ without being $0$, thus $\sum_{r=1}^{\infty} d(r) a_r / t^r$ must be irrational.

If we instead ask when $\sum_{r=1}^{\infty} d(r)(-1)^r a_r / t^r$ is irrational, the answer is not quite as satisfying. In order to make certain that the tail of the series does not vanish completely, in our proof above, we wanted to have some large prime $p$ and consider $d(p)(-1)^p / t^p$. If the corresponding digit $a_p$ is $0$ then this method of attack fails. So, we know $\sum_{r=1}^{\infty} d(r)(-1)^r a_r / t^r$ will be irrational if for all sufficiently large primes, $a_p \neq 0$.

We will now show that $d(r)$ has a similar irrationality-creating property in certain generalized Luroth series expansions. Our definitions of the GLS expansion come predominantly from the work of Dajani and Kraaikamp.

Consider a partition $\mathcal{I}=\left\{ (l_n,r_n]: n \in \mathbb{D}\right\}$ of $(0,1]$ where $\mathbb{D} \subset \mathbb{N}$ and $\sum_{n \in \mathbb{D}} (r_n -l_n) = 1$. For each $n \in \mathbb{D}$, let $I_n = (l_n,r_n]$.

Every real number in $(0,1]$ corresponds to a unique expansion of the form $$\sum_{r=1}^{\infty} h_r \prod_{i=1}^{r-1} s_i$$ where $h_r,s_r$ are the left endpoint and interval length of some $I_n$ respectively. This is known as a GLS($\mathcal{I}$) expansion.

We call a partition $\mathcal{I}$, as defined above, regular if each $l_n,r_n$ is rational, and when put in lowest terms the denominator of $l_n$ divides the denominator of $r_n-l_n$. Further, call $\mathcal{I}$ $t$-regular if in addition to being regular, the denominator of $r_n-l_n$ (in lowest terms) is divisible by $t$ for all $n$.

If $\mathcal{I}$ is a $t$-regular partition where the numerator of each $r_n-l_n$ is $1$, then for all GLS($\mathcal{I}$) expansions $$\sum_{r=1}^{\infty} h_r \prod_{i=1}^{r-1} s_i$$ we have that $$\sum_{r=1}^{\infty} d(r) h_r \prod_{i=1}^{r-1} s_i$$ is irrational.

If $\mathcal{I}$ is a $t$-regular partition where the numerator of each $r_n-l_n$ is relatively prime to $t$ and for each prime $p$ dividing a numerator, we have $$\sum_{p^k || (r_n-l_n)} k*(r_n-l_n) <0$$ then for lebesgue-almost all GLS($\mathcal{I}$) expansions $$\sum_{r=1}^{\infty} h_r \prod_{i=1}^{r-1} s_i$$ we have that $$\sum_{r=1}^{\infty} d(r) h_r \prod_{i=1}^{r-1} s_i$$ is irrational.

Provided no $l_n=0$ the above results also hold if $d(r)$ is replaced with $d(r)(-1)^r$.

In the first case, note that since $s_r$ always takes the form of $1$ divided by some positive multiple of $t$ that we can write $h_r \prod_{i=1}^{r-1} s_i$ as $a_r/b_r$ with integers $a_r,b_r$, where $a_r$ is the numerator of $h_r$, $b_r|b_{r+1}$ and $t^{r-1}|b_r$.

For the second case, it suffices to consider those numbers normal in the GLS($\mathcal{I}$) expansion, that is, those expansions $\sum_{r=1}^{\infty} h_r \prod_{i=1}^{r-1} s_i$ where $$\lim_{r \to \infty} \frac{1}{r} \#\{ h_r = l_n\}$$ is well-defined and equals $r_n-l_n$. Our condition $\sum_{p^k || (r_n-l_n)} k*(r_n-l_n) <0$ implies that for all sufficiently large $r$ we can write $h_r \prod_{i=1}^{r-1} s_i$ as $a_r/b_r$ with integers $a_r,b_r$, where $a_r$ is the numerator of $h_r$, $b_r|b_{r+1}$ and $t^{r-1}|b_r$.

Acknowledgements
================

The author acknowledges support from National Science Foundation grant DMS 08-38434 “EMSW21-MCTP: Research Experience for Graduate Students.” The author would also like to thank Paul Pollack and Paul Spiegelhalter for their assistance.

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[^1]: Here we switch back to the convention that finite expansions are assumed to end on an infinite string of zeros.
