Erdős Problem 1051
Statement and result.
Let $a_1<a_2<\cdots$ be positive integers and suppose $\liminf_{n\to\infty}a_n^{1/2^n}>1$. Then \[
S=\sum_{n=1}^{\infty}\frac1{a_na_{n+1}}
\] is irrational. We give an elementary proof under exactly this stated hypothesis.
Growth and record indices.
There are $c>0$ and $n_0$ such that $\log a_n\geq c2^n$ for $n\geq n_0$. Fix a real number \[
\frac{1+\sqrt5}{2}<\gamma<2.
\] The sequence $b_i=(\log a_i)/\gamma^i$ is unbounded, so it has record indices $j$ tending to infinity, with $b_i\leq b_j$ for every $i\leq j$. At such an index, put $A=a_j$. Then \[
a_i\leq A^{\gamma^{i-j}}\quad(1\leq i\leq j).
\tag{1}
\] In particular, \[
\prod_{i=1}^{j-2}a_i\leq A^{1/(\gamma(\gamma-1))},
\qquad
\prod_{i=1}^{j-1}a_i\leq A^{1/(\gamma-1)}.
\tag{2}
\] These follow by extending the relevant finite geometric sums to infinite sums.
A uniform bound for the far tail.
For sufficiently large $j$, let \[
J=\left\lceil\log_2\!\left(\frac{\log A}{c}\right)\right\rceil.
\] Since $\log A\geq c2^j$, we have $J\geq j$. Monotonicity gives $a_n\geq A$ for $n\geq j$, while for $r\geq1$, \[
a_{J+r}\geq\exp(c2^{J+r})\geq A^{2^r}.
\] Consequently, once $A\geq2$, \[
\sum_{n=j}^{\infty}\frac1{a_na_{n+1}}
\leq\sum_{n=j}^{\infty}\frac1{a_n^2}
\leq\frac{J+1}{A^2}+\sum_{r=1}^{\infty}A^{-2^{r+1}}
\leq\frac{J+2}{A^2}.
\tag{3}
\] Here $J=O_c(1+\log\log A)$, and the last geometric comparison is deliberately loose.
An integer that would tend to zero.
Suppose $S=p/q$ with integers $p$ and $q\geq1$. For a record index $j\geq3$, set \[
D_j=\prod_{i=1}^{j-1}a_i,\qquad
S_{j-2}=\sum_{i=1}^{j-2}\frac1{a_i a_{i+1}}.
\] Every denominator in $S_{j-2}$ divides $D_j$; hence $qD_j(S-S_{j-2})$ is a positive integer. On the other hand, separating off its first tail term and using $2$--$3$, we obtain \[
\begin{split}
0<D_j(S-S_{j-2})
&\leq
\frac{\prod_{i=1}^{j-2}a_i}{A}
+\frac{(J+2)D_j}{A^2}\\
&\leq
A^{1/(\gamma(\gamma-1))-1}
+(J+2)A^{1/(\gamma-1)-2}.
\end{split}
\] Both exponents are negative: the first because $\gamma(\gamma-1)>1$, and the second because $\gamma>3/2$. As $A\to\infty$ through record indices, the right side tends to zero, even with the $O(\log\log A)$ factor. This contradicts the positive-integer lower bound $qD_j(S-S_{j-2})\geq1$, and proves irrationality.
Scope and attribution.
The proof establishes the stated sufficient condition without imposing divisibility on the $a_n$. It does not assert irrationality for every increasing sequence: for example $a_n=n$ gives a telescoping rational sum. Barreto, Kang, Kim, Kovač and Zhang prove substantially stronger irrationality criteria in their 2026 work, including a threshold involving the golden ratio; their paper attributes the original resolution to the Aletheia research agent. Their theorem is background here, not an input to the argument above. Sources checked 24 September 2026:
https://www.erdosproblems.com/1051 and
https://arxiv.org/abs/2601.21442. No novelty or local formal verification is claimed.
Completion estimate.
COMPLETION: 100%