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Use the counting and probability notation. For every integer $n\ge1$,

$$\frac{R_n}{2^n} =\mathbb P(Z_n\le2-H_n) =\mathbb P(Z_n\ge H_n-2). \tag{1}$$

For every $x>0$ and $t>0$,

$$\mathbb P(Z_n\ge t) \le e^{-xt}\mathbb E e^{xZ_n} =e^{-xt}\prod_{i=1}^n\cosh(x/i). \tag{2}$$

The identity (1) holds even when $H_n\le2$. To use (2) with $t=H_n-2$, we require $H_n>2$, which holds for all sufficiently large $n$.

Proof. Write $\delta_i=\mathbf1_{i\in S}=(1+\varepsilon_i)/2$. Then

$$\sum_{i=1}^n\frac{\delta_i}{i} =\frac{H_n+Z_n}{2}.$$

Thus the reciprocal sum is at most one exactly when $Z_n\le2-H_n$. The subset-to-sign map is a bijection. Each sign vector has probability $2^{-n}$. This proves the first equality in (1). The bijection $(\varepsilon_i)\mapsto(-\varepsilon_i)$ sends this event to $Z_n\ge H_n-2$, proving the second equality.

For a sign vector with $Z_n\ge t$, the number $e^{x(Z_n-t)}$ is at least one. For every other vector it is positive. Averaging the pointwise inequality $\mathbf1_{Z_n\ge t}\le e^{x(Z_n-t)}$ over the finite probability space gives the first inequality in (2). Independence gives

$$\mathbb E e^{xZ_n} =\prod_{i=1}^n\mathbb E e^{x\varepsilon_i/i} =\prod_{i=1}^n\frac{e^{x/i}+e^{-x/i}}2,$$

which is the product in (2). This proves the exponential-moment estimate without importing a separate deviation theorem. $\square$

When $H_n>2$, the two tail events in (1) are disjoint. Consequently the absolute tail $\{|Z_n|\ge H_n-2\}$ has probability $2R_n/2^n$, not $R_n/2^n$. The source correctly uses a lower tail and then an upper tail by symmetry.

Source. Steinerberger, arXiv:2403.17041v5, p. 2, §§2.1–2.2. This is the complete finite argument from those sections, with the relation between the two events made explicit.

Bears on. #297 only through the Theorem, whose proof uses it.