Let $n \ge 1$. In the lecture we wrote that
$$2\exp\left( -\frac{n \log n}{2 \left(\frac{n-1}{4}+\frac{\sqrt{n \log n}}{3} \right)} \right) = 2\exp\left(-2(1+\mathcal{o}(1)) \log n\right).$$
I know that $\exp(-\cdot)$ is strictly decreasing, so I suppose that we need to argue that for any $\varepsilon > 0$ holds
$$\frac{n}{2\left(\frac{n-1}{4} + \frac{\sqrt{n \log(n)}}{3}\right)} > 2+\varepsilon \quad \text{ as $n \rightarrow \infty$}.$$
But how to go on from here?
You have to show that $\frac{n}{2\left(\frac{n-1}{4} + \frac{\sqrt{n \log(n)}}{3}\right)} = 2 + o(1).$ Let $\varepsilon > 0$ be fixed. Writing $$\frac{n}{2\left(\frac{n-1}{4} + \frac{\sqrt{n \log(n)}}{3}\right)} = \frac{2}{\frac{n - 1}{n} + \frac{1}{12} \sqrt{\frac{\log n}{n}}} =: R_n$$ and noting that the denominator converges to $1$ gives you the existence of an $n_\varepsilon \in \mathbb{N}$ such that for all $n \geq n_\varepsilon,$ one has that $\vert \frac{n - 1}{n} + \frac{1}{12} \sqrt{\frac{\log n}{n}} - 1 \vert \leq 1 - \frac{2}{2 + \varepsilon} =: \delta,$ so that $R_n \leq \frac{2}{1 - \delta} = 2 + \varepsilon.$ I hope this helps. :)