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 $$\lim_{k\to \infty }\frac{a_{k+1}}{a_k}=1.$$

This isn't natural number-valued, but just taking the ceiling should make everything work. Take $$a_k = e^{k/\log k},$$
then $$\frac{1}{(\log (a_k))^{1+\delta}}= \left( \frac{\log k}{k} \right)^{1+\delta},$$ whose sum converges. Notice that $$\begin{align} \log \left( \frac{a_{k+1}}{a_k} \right) &= \frac{k+1}{\log(k+1)}-\frac{k}{\log k} \\
 &= k\left(\frac{1}{\log(k+1)} - \frac{1}{\log k} \right) + \frac{1}{\log(k+1)}, \end{align}$$
then as the derivative of $\frac{1}{\log x}$ is $-\frac{1}{x (\log x)^2}$, the first term converges to zero using the mean value theorem to show that it's bounded by something like $\frac{1}{(\log k)^2}$ and the second also converges to zero, implying $a_{k+1}/a_k$ converges to $1$.
