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This is a complete author-recorded reconstruction by Plasma AI of the increasing-block idea recorded on the Price claim page. That claim retains its credit and standing; no novelty, independent review, or examination of its manuscript or formal proof is asserted here.

There exists a strictly increasing sequence of positive integers $a_n$, indexed from $n=1$, with $\liminf a_n^{1/n}=1$ such that every positive-integer sequence $b_n$ with $b_n/a_n\to1$ has a convergent irrational reciprocal sum. The argument below addresses this necessary-growth subquestion of Problem 263, not its fixed sequence $2^{2^n}$.

For each $k\ge 1$, choose an integer $M_k\ge 2$ and put $L_k=\lceil k\log_2 M_k\rceil$. The sequence $a_n$ is the increasing concatenation of the blocks

$$M_k,M_k+1,\ldots,M_k+L_k-1.$$

Write $N_k=\sum_{j\le k}L_j$ and $P_k=\prod_{n\le N_k}(2a_n)$, with $P_0=1$. Choose $M_k$ recursively so that its block lies strictly beyond the preceding block, $L_k\le M_k$, and

$$\frac{L_k}{M_k}\leq\frac{2^{-k}}{P_{k-1}}.$$

This is possible: at the time of choosing $M_k$, $k$ and $P_{k-1}$ are fixed, while $\lceil k\log_2 M\rceil/M\to 0$. The blocks are finite and disjoint, so this defines a strictly increasing sequence of positive integers. Also $\sum 1/a_n \le \sum L_k/M_k \le \sum 2^{-k}<\infty$.

Let any positive-integer sequence $b_n$ satisfy $b_n/a_n\to 1$. Eventually $a_n/2\le b_n\le 2a_n$, and its reciprocal series converges. Absorb the finitely many early upper-bound exceptions into a constant $C>0$. At block endpoints, its actual denominator product obeys

$$B_{N_k}:=\prod_{n\leq N_k}b_n\leq C P_k.$$

For all large $k$, the positive tail $T_k=\sum_{n>N_k}1/b_n$ satisfies

$$0<T_k\leq2\sum_{j>k}\frac{L_j}{M_j} \leq\frac{2}{P_k}\sum_{j>k}2^{-j} =\frac{2^{1-k}}{P_k},$$

because $P_{j-1}\ge P_k$. If the complete sum were the rational $u/v$ with integer $v>0$, then $v B_{N_k} T_k$ would be a positive integer: multiplication by the product clears every denominator in the prefix. But it is at most $v C 2^{1-k}$, eventually less than 1, a contradiction. Thus this one increasing sequence works for every such perturbation.

At block endpoints $a_{N_k}<M_k+L_k\le 2M_k$ and $N_k\ge L_k\ge k\log_2 M_k$, so

$$0\leq\frac{\log a_{N_k}}{N_k} \leq\frac{\log(2M_k)}{k\log_2M_k} \leq\frac{2\log2}{k}\longrightarrow0.$$

Since all $a_n\ge 1$, it follows that $\liminf a_n^{1/n}=1$. Consequently the proposed necessary limit to infinity is false even for strictly increasing sequences. This proves the stated growth counterexample. The fixed sequence $2^{2^n}$ is not addressed.