Let $R > 1$ and $\lambda \in \mathbb{R}$ be such that $$ \int_{1}^R \mathrm{tanh}(\frac{\lambda x}{2}) \frac{d x}{x} = \log R -2. $$ Then standard techniques in large deviation theory yield $$ \frac{1}{n} \log | \{ I \subseteq [n R^{-1},n] | \sum_{i \in I} \frac{1}{i} \leq 1 \} | \longrightarrow \int_{1}^R \phi(\lambda x)\frac{d x}{x^2} $$ where $\phi(u) = \log2 +\log \cosh (\frac{u}{2}) - \frac{u}{2} = \log(1 + e^{-u})$. For $R = e^2$ one has $\lambda =0$ and the limit is $(1 - e^{-2})\log 2$ : this recovers Lucia's first result.

Moreover, letting $R$ tend to infinity, then $\lambda$ tends to Lucia's constant $\alpha$, and the limit integral matches Lucia's upper bound. In particular Lucia's "second guess" was correct.

Let $R > 1$ and $\lambda \in \mathbb{R}$ be such that $$ \int_{1}^R \mathrm{tanh}(\frac{\lambda x}{2}) \frac{d x}{x} = \log R -2. $$ Then standard techniques in large deviation theory yield $$ \frac{1}{n} \log | \{ I \subseteq [n R^{-1},n] | \sum_{i \in I} \frac{1}{i} \leq 1 \} | \longrightarrow \int_{1}^R \phi(\lambda x)\frac{d x}{x^2} $$ where $\phi(u) = \log2 +\log \cosh (\frac{u}{2}) - \frac{u}{2} = \log(1 + e^{-u})$. For $R = e^2$ one has $\lambda =0$ and the limit is $(1 - e^{-2})\log 2$ : this recovers Lucia's first result.

Moreover, letting $R$ tend to infinity, then $\lambda$ tends to Lucia's constant $\alpha$, and the limit integral matches Lucia's upper bound. In particular Lucia's "second guess" was correct.

EDIT: Some details about the said "standard techniques". Let $(X_i)_i$ be independent Bernoulli random variable ($= \pm 1$ with probability $1/2$), and let $(t_i)_{i=1}^n$ be positive real numbers lying in a fixed interval $(-R,R)$. We are going to study the probability that $S = \sum_{i=1}^n t_i X_i$ is $\geq c$, for some $c \in \mathbb{R}$. Let $\lambda \in \mathbb{R}$ be such that $$ \frac{1}{n} \sum_{i=1}^n t_i \tanh(\lambda t_i) = c. $$ This amounts to say that $E[\frac{1}{n} S e^{\lambda S}] = c E[e^{\lambda S}]$. Such a $\lambda$ exists as soon as $|c| < \frac{1}{n} \sum_{i=1}^n t_i$. One checks that $$ E[(\frac{1}{n} S)^2 e^{\lambda S}]E[e^{\lambda S}]^{-1} = c^2 + \frac{1}{n^2} \sum_{i=1}^n t_i^2(1- \tanh(\lambda t_i)^2) $$ so that the variance of $\frac{1}{n} S$ w.r.t. the measure weighted by $e^{\lambda S} E[e^{\lambda S}]^{-1}$, is at most $R^2 n^{-1}$. In particular $$ E[ \mathbb{1}_{\frac{1}{n} S \in [c,c+\epsilon]} e^{\lambda S}] E[e^{\lambda S}]^{-1} = \frac{1}{2} + O(R^2 \epsilon^{-2} n^{-1}). $$ Thus $$ P( S \geq nc) \geq e^{-n \lambda (c+\epsilon)}E[ \mathbb{1}_{\frac{1}{n} S \in [c,c+\epsilon]} e^{\lambda S}] = e^{-n \lambda (c+\epsilon)} E[e^{\lambda S}] \left( \frac{1}{2} + O(R^2 \epsilon^{-2} n^{-1}) \right) $$ On the other hand $$ P( S \geq nc) \leq e^{-n \lambda c} E[e^{\lambda S}] $$ so that $$ \frac{1}{n} \log P( S \geq nc) = - \lambda c + \frac{1}{n} \log E[e^{\lambda S}] + o_{R}(1) $$ For the application above, just use $t_i = \frac{n}{2i}$ (indexed by $i$ between $n R^{-1}$ and $n$) and $c = \frac{1}{2} \sum_{n R^{-1} < j \leq n} \frac{1}{j} - 1$.