In my lecture notes, the following is written:
$$1- (1-1/n)^{\varepsilon n} \leq \varepsilon + \mathcal{O}(\varepsilon^2)$$
as $\varepsilon \rightarrow 0$ and $n$ some fixed constant (non-negative integer), but I do not understand why.
Everything I get is
$$1-(1-1/n)^{\varepsilon n} \leq e^{-(1-1/n)^{\varepsilon n}}$$
but I am not able to get to the statement.
Thank you for your help!
The power series for an analytic function is an asymptotic series as well, so we can say that $\exp \delta \sim 1+\delta + \mathcal O(\delta^2)$ as $\delta \to 0$.
Applying this to your expression yields $$ 1 - (1-1/n)^{\epsilon n} = 1 - \exp(\epsilon n \log(1-1/n)) \sim -\epsilon n \log (1-1/n) + \mathcal O(\epsilon^2)~~. $$
Note that $-n\log(1-1/n) > 1$ for each $n>1$, and in fact tends to 1 as $n\to\infty$. In particular, if $n$ is constant, this is also a constant. I'm not clear on which definition of asymptotic notation you're using, but perhaps this information is enough to conclude the desired result?