I am trying to reconstruct the proof that Erdös-Renyi-Graphs with an edge probabiblity 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 a second question regarding the same proof.
In this step, it is claimed without further explanation that for $n \to \infty$
$$ \sum_{k = n^{\lceil3/4\rceil} + 1}^{n/2} (e/n^{1/4})^k \leq n^{-n^{3/4}/5} $$
I see how you can say $e < n$, so $e/n^{1/4} \leq n^{-3/4}$ but that's about it. I have no clue how you get rid of the series, how $n^{-3/4}$ suddenly becomes an exponent or where the magical $/5$ comes from.
Context
This is only one half of the last inequality in Jackson's book. The whole last inequality in his book reads:
$$\sum_{k = 2}^{n^{3/4}} e^{k(1-f(n))}k^{-k}e^{2k^2 \ln n / n} + \sum_{k=n^{3/4}}^{n/2} \left(\frac{en}{k}\right)^k e^{-knp/2} \leq 3e^{-f(n)} + n^{-n^{3/4}/5}$$
However, Bollobas treats both sums on the left hand side separately, and takes the intermediate step of showing that $\left(\frac{en}{k}\right)^k e^{-knp /2} \leq (en^{1/4})^kn^{-k/2} \leq (e/n^{1/4})^k$. These steps I understand, it's only how to get to $n^{-n^{3/4}/5}$ from there.
[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.
First we will prove that $$\displaystyle n\left(\dfrac{e^{n^{3/4}}}{n^{\frac{n^{3/4}}{4}}}\right)\le n^{-\frac{n^{3/4}}{5}}\Longleftrightarrow \dfrac{e^{n^{3/4}}}{n^{\frac{n^{3/4}}{4}}}\cdot n^{\left(\frac{n^3/4}{5}+1\right)}\le 1\Longleftrightarrow n^{\frac{n^{3/4}}{20}-1}\ge e^{n^{3/4}}$$ Which is true when $n> e^{21}$, since $\displaystyle n^{\frac{n^{3/4}}{20}-1}>e^{21\left(\frac{n^{3/4}}{20}-1\right)}$ and, $e^{21\cdot \frac{3}{4}}>21\cdot 20$ for sure so that $21\cdot\dfrac{n^{3/4}}{20}-21>n^{3/4}\Longleftrightarrow n^{3/4}>21\cdot 20$, and thus $\displaystyle e^{21\left(\frac{n^{3/4}}{20}-1\right)}\ge e^{n^{3/4}}$ as desired.
Then, by the original inequality, since $e/n^{1/4}<1$ for sufficiently large $n$ so $(e/n^{1/4})^p<(e/n^{1/4})^{q}$ for any $p>q$ that is, $$\sum_{k=n^{3/4}}^{n/2}\left(\dfrac{e}{n^{1/4}}\right)^{k}\le n\left(\dfrac{e^{n^{3/4}}}{n^{\frac{n^{3/4}}{4}}}\right)\le n^{-\frac{n^{3/4}}{5}}$$ as desired, from the preceded inequality $\Box$