Why is $\sum_{k=1}^{n^{3/4}} e^{(1-\omega(n))k}k^{-k}e^{2k^2(\log n)/n} \leq 3e^{-\omega(n)}$?

Question

I am trying to reconstruct the proof that Erdös-Renyi-Graphs with an edge probability above $\ln n / n$ are connected w.h.p. I fail to see the last step. The proof in its whole is from Section 4.2.3 of the book by Matthew O. Jackson [1], but this last part is taken from a book by Bollobas [2] (page 234, equations 23 and 24). I can't find a freely available PDF of the Bollobas book, so under "Context" below I outline the steps by Bollobas which Jackson omits. I also currently ask another question regarding the same proof.

In this step, it is claimed without further explanation that for $n \to \infty$ and some function $\omega(n)$ with $\omega(n) \to \infty$, $\omega(n) < \ln n$, it holds that

$$\sum_{k=1}^{n^{3/4}} e^{(1-\omega(n))k}k^{-k}e^{2k^2(\log n)/n} \leq 3e^{-\omega(n)}$$

Here is my best attempt so far:

• $k^{-k} \leq 1$ for $k \geq 1$, so we drop $k^{-k}$ from the product.
• $e^{(1-\omega(n))k}$ becomes maximal for $k = 1$, since we may assume $\omega(n) > 1$
• $e^2k^2(\log n)/n$ becomes maximal for $k = n^{3/4}$

Thus, we now are at:

$$\sum_{k=1}^{n^{3/4}} e^{(1-\omega(n))}e^{2(n^{3/4})^2(\log n)/n} = \sum_{k=1}^{n^{3/4}} e \cdot e^{-\omega(n)}e^{2(n^{9/16})(\log n)/n} = \sum_{k=1}^{n^{3/4}} e \cdot e^{-\omega(n)}e^{2(n^{-7/16})(\log n)}$$

Now the sum is independent of $k$, so we can upper-bound this as

$$\leq n^{3/4} \cdot e \cdot e^{-\omega(n)}e^{2(n^{-7/16})(\log n)} $$

For $n \to \infty$, it holds that $2(n^{-7/16})(\log n) \to 0$, so

$$\leq n^{3/4} \cdot e \cdot e^{-\omega(n)} $$

This is where I am stuck. I don't see how to get rid of the $n$. So, maybe my attempt of resolving that sum was wrong. Obviously, for $\omega(n) \to \infty$, $3e^{-\omega(n)} \to 0$, which is what the proof ultimately tries to show. However, since $\omega(n) \leq \ln n$, it holds that $n^{3/4} \cdot e \cdot e^{-\omega(n)} \not\to 0$, and I must be missing some trick.

Context

This is only one half of the last inequality in Jackson's book. The whole last inequality in his book reads:

$$\sum_{k = 2}^{n^{3/4}} e^{k(1-\omega(n))}k^{-k}e^{2k^2 \ln n / n} + \sum_{k=n^{3/4}}^{n/2} \left(\frac{en}{k}\right)^k e^{-knp/2} \leq 3e^{-f(n)} + n^{-n^{3/4}/5}$$

However, Bollobas treats both sums on the left hand side separately. See the other question linked above for the other half.

[1] Jackson, Matthew O. Social and economic networks. Princeton University Press, 2010.

[2] Bollobás, Béla. Modern graph theory. Vol. 184. Springer Science & Business Media, 2013.