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In the growing range of Lemma 2, set $q=cn$, $p_m=(1+e^{q/m})^{-1}$, and let $Y_m$ be independent Bernoulli variables of means $p_m$. Put $Z=\sum_mY_m/m$, so $\mathbb EZ=x$. Uniformly there,

$$\operatorname{Var}Z\asymp_{x_0}q^{-1},\qquad \sum_m\mathbb E|(Y_m-p_m)/m|^3=O_{x_0}(q^{-2}).$$

The same estimates hold after removing any one summand; a deterministic shift does not affect them. Berry–Esseen consequently gives normal-approximation error $O_{x_0}(q^{-1/2})$ in either case.

Source: published PDF, p. 5, equations (3)–(5) and the coordinate calculation. Integral/variation bounds replace the source's per-slab lower estimates, which can have empty integer intervals. Absolute third moments, as printed in the published version, are required.

Bears on. Problem 297.

Proof

For fixed $r\ge1$ let $f_r(u)=e^{-q/u}/u^r$ for $u>0$, with $f_r(0)=0$. It is unimodal, with maximum $O_r(q^{-r})$ and total variation of that order. The unit-interval Riemann comparison therefore gives

$$\sum_{m=1}^n\frac{e^{-q/m}}{m^r} =q^{1-r}\int_c^\infty e^{-t}t^{r-2}\,dt+O_r(q^{-r}). \tag{1}$$

For $r=2,3$, $c\le C(x_0)$ bounds the integral above and below by positive constants depending only on $x_0,r$. Since $q\to\infty$ uniformly, the error is smaller than the main term. Thus these sums are $\Theta_{x_0,r}(q^{1-r})$. For later use, $r=1$ instead gives the upper bound $O(1+\log_+(1/c))$, because the integral has only a logarithmic divergence at zero and $q^{-1}\le1$ eventually.

For $t=q/m>0$,

$$\frac14e^{-t}\le p_m(1-p_m)=\frac{e^{-t}}{(1+e^{-t})^2} \le e^{-t}.$$

Sum this divided by $m^2$ and use (1) for the variance. For a Bernoulli variable of mean $p$,

$$\mathbb E|Y-p|^3 =p(1-p)\bigl((1-p)^2+p^2\bigr)\le p\le e^{-q/m}.$$

Equation (1) with $r=3$ proves the third-moment bound. The variance of one summand is at most $\sup_{u>0}e^{-q/u}/u^2=O(q^{-2})$. Removing it from a variance bounded below by a positive multiple of $q^{-1}$ preserves that lower bound for large $q$, uniformly in the index. The third-moment upper bound can only decrease. Independence remains. Substitute these estimates into Lemma 3 to get error $O(q^{-2}/(q^{-1})^{3/2})=O(q^{-1/2})$.