I am checking the classic paper by Erdős and Rényi, "On the existence of a factor of degree one of a connected random graph" with the link here. I am curious about the computation of the upper bound in the 6th set, the first line on page 366 (page 8 of the pdf). In particular, how to use the inequalities in (1.1) - (1.10) to get the following bound magically? $$ \sum_{r<\frac{8n\log\log^2n}{\log^2n},} \sum_{r+1\le k<\frac{8n\log\log^2n}{\log^2n}}\frac{n^{k+r}}{k!r!}\sum_{r+8\le s\le kr} \binom{kr}{s}\Big(\frac{\log n}{n}\Big)^s \cdot \frac{1}{n^k} = O(n^{e^2-8}). $$ It might be easy to figure out the term $n^{-8}$. But I wonder if there is any elegant way to get $n^{e^2}$. (There was a typo in the linked file. It should be $r<\frac{8n\log\log^2n}{\log^2n}$ but not $r<\frac{8\log\log^2n}{\log^2n}$.)
I attached inequalities (1.1)–(1.10) here.
$$ \binom n r \le \frac{n^r}{r!}\quad(r = 1, 2, \dotsc; n = r, r + 1, \dotsc). \label{482430_1.1}\tag{1.1} $$ $$ \frac{x - a}{y - a} \le \frac x y\quad\text{for}\quad 0 < a \le x \le y. \label{482430_1.2}\tag{1.2} \\ $$ If $0 \le b - a \le B - A$, $0 \le a \le b \le B, 0 \le A$, $$ \frac{\binom{B - A}{b - a}}{\binom B b} \le \left(1 - \frac{A - a}{B - a}\right)^{b - a}\left(\frac b B\right)^a. \label{482430_1.3}\tag{1.3} $$ $$ 1 - x \le e^{-x}\quad\text{if}\quad x \ge 0. \label{482430_1.4}\tag{1.4} $$ $$ n! \ge \left(\frac n e\right)^n\quad\text{for}\quad n \ge 1. \label{482430_1.5}\tag{1.5} $$ If $\lambda > 1$, $0 < \delta < \frac1{\lambda e}$ we have $$ \sum_{n\delta\lambda e \le r \le n} \binom n r\delta^r = O\left(\frac1{\lambda^{\delta\lambda e n}}\right). \label{482430_1.6}\tag{1.6} $$ For $0 < p < \tfrac1 2$ and $h(p) = p\log \frac1 p + (1 - p)\log \frac1{1 - p}$, one has $$ \sum_{k \le n p} \binom n k = O(\sqrt n\cdot e^{n h(p)} ).\label{482430_1.7}\tag{1.7} $$ If $0 < \alpha < 1$, $c > 0$ $$ \sum_{\frac{n c}{\log n} \le k} \binom n k\frac1{n^{\alpha k}} = e^{-\alpha c n + o(n)}. \label{482430_1.8}\tag{1.8} $$ $$ 1 - x + \frac{x^2}2 \le e^{-x + \frac{x^3}3}\quad\text{if}\quad 0 \le x < 1. \label{482430_1.9}\tag{1.9} $$ $$ \binom n k \le \frac{n^k}{k!} e^{-\frac{k(k - 1)}{2n}}\quad\text{if}\quad 0 < k \le n. \label{482430_1.10}\tag{1.10} $$
