Select a subsequence to obtain a convergent series. Does there exists strictly increasing sequence  $\{a_k\}_{k\in\mathbb N}\subset\mathbb N$,
such that
$$
\sum_{k=1}^{\infty}\frac{1}{(\log a_k)^{1+\delta}}\lt \infty,
$$
where $\delta>0$ given and $$\lim_{k\to \infty }\frac{a_{k+1}}{a_k}=1.$$

Answer. Try
$$
a_k=\left\lfloor 2^{k^{\color{red}{1/(1+\delta/2)}}}\right\rfloor.
$$ 
Then $\dfrac{a_{k+1}}{a_k}\to 1$, and 
$$
\frac{1}{(\log a_k)^{1+\delta}}\approx\frac{c}{n^{1+\delta/2}}.
$$
