Is $\exp (-\alpha\log (n)^\beta(1+o(1)))=o(1)$?

Question

Referring to Butucea, $\hat s_n$ is an estimator of $s_k$.

In (5) on p.902 there is the statement that $$\mathrm{P}\left(\hat{s}_n \neq s_k\right) \leq \exp \left(-\frac{A^2}{4} 2^{2 \beta^{\prime} / \bar{s}}(\log n)^{2\left(\delta-\beta^{\prime}\right) / \bar{s}}(1+o(1))\right)\quad \forall k\in\{1,\dots,N\}$$ where $A\in\mathbb R^+,\beta^\prime\in\mathbb R^+$, $\bar s\in(0,2]$, $\delta>\beta^\prime$.

I think this statement is made to show that $\hat s_n$ is a consistent estimator of $s_k$, i.e.

$$\forall\varepsilon>0:\lim_{n\to\infty}\mathrm P(|\hat s_n-s_k| > \varepsilon) = 0.$$ To show consistency one could calculate like this:

$$\mathbb P(|\hat s_n-s_k| > \varepsilon)\le\mathbb P(|\hat s_n-s_k| > 0)=\mathbb P(\hat{s}_n \neq s_k)\overset!=o(1).$$

So we have to argue that $$\exp \left(-\frac{A^2}{4} 2^{2 \beta^{\prime} / \bar{s}}\log (n)^{2\left(\delta-\beta^{\prime}\right) / \bar{s}}(1+o(1))\right)=o(1).$$

Renaming the constants we have that for $\alpha,\beta>0$ must hold $$\exp \left(-\alpha\log (n)^\beta(1+o(1))\right)=o(1).$$

Here I am not really sure why this should be true for $n\to\infty$.

Answer 1

As $n\to\infty$, $\log(n)^{\beta}\to\infty$ and consequently $-\alpha\log(n)^{\beta}\to-\infty$. Since $1+o(1)\to 1$ as $n\to\infty$, we get

$$\lim_{n\to\infty}-\alpha\log(n)^{\beta}(1+o(1))=-\infty\cdot 1=-\infty$$ $$\implies\lim_{n\to\infty}\exp\left(-\alpha\log(n)^{\beta}(1+o(1))\right)=\exp(-\infty)=0$$