Another conditionl leading to irrationality of $\sum _{k=1}^ \infty \dfrac 1{n_k}$? If $\{n_k\}$ is a strictly increasing sequence of positive integers such that $\lim \inf _{k \to \infty} n_k ^{1/2^k} >1$ and $\lim _{k \to \infty} n_k^{1/2^k}$ does not exist , then is it true that $\sum _{k=1}^ \infty \dfrac 1{n_k}$ is irrational ?
 A: If we assume an additional condition, then the statement is true ; note that if $\left\{n_{k}\right\}$  is a strictly increasing sequence of positive integers  satisfying the hypothesis,  $\left\{n_{k}\right\}$ is not of the following form: 

$\left\{n_{k}\right\}$ is a sequence such that for some $N\in\mathbb{N}$, $n_k\leq 2^k$ when $k\geq N$. 

In this case, we have: 
\begin{eqnarray*}
 \lim_{k\rightarrow\infty}\ln\left(n_{k}^{1/2^{k}}\right)
 & = & \lim_{k\rightarrow\infty}\frac{\ln\left(n_{k}\right)}{2^{k}} \\
  &\leq& \lim_{k\rightarrow\infty}\frac{\ln\left(2^{k}\right)}{2^{k}}=\lim_{k\rightarrow\infty}\frac{k\ln\left(2\right)}{2^{k}}=0
\end{eqnarray*}
Then, $\lim_{k\rightarrow\infty}n_{k}^{1/2^{k}}=1$.
Furthermore,  $\left\{n_{k}\right\}$ is not of the following form:

There is $\left\{n_{k_i}\right\}$ a subsequence of $\left\{n_{k}\right\}$ such that for some $N\in\mathbb{N}$, $n_{k_{i}}\leq 2^k_{i}$ when $i\geq N$. 

In this case, proceeding as in the previous case we have $\lim_{i\rightarrow\infty}n_{k_i}^{1/2^{k_i}}=1$, therefore, $\liminf_{k\rightarrow\infty}n_{k}^{1/2^{k}}\leq 1$.
Therefore, $\left\{n_{k}\right\}$ must be of the form:

$\left\{n_{k}\right\}$ is a sequence such that for some $N\in\mathbb{N}$, $n_k\geq 2^k$ when $k\geq N$. 

But not all this sequences are satisfying the hypothesis, note that if $\left\{n_{k}\right\}$ is one of this sequence such that there is a subsequence $\left\{n_{k_{i}}\right\}$ such that $$n_{k_{i}+1}\leq n_{k_{i}}^{2}-n_{k_{i}}+1$$ for all  $i$, then
\begin{eqnarray*}
\lim_{i\rightarrow\infty}\ln\left(n_{k_{i}+1}^{1/2^{k_{i}+1}}\right)
 & = & \lim_{i\rightarrow\infty}\frac{\ln\left(n_{k_{i}+1}\right)}{2^{k_{i}+1}} \\
  &\leq& \lim_{i\rightarrow\infty}\frac{\ln\left(n_{k_{i}}^{2}-n_{k_{i}}+1\right)}{2^{k_{i}}}&\leq&\lim_{i\rightarrow\infty}\frac{\ln\left(2^{\frac{n_{k_{i}}}{2^{n_{k_{i}}}}}\right)}{2^{k_{i}+1}} \\ &=& \lim_{i\rightarrow\infty}\frac{n_{k_{i}} \ln\left(2\right) }{2^{n_{k_{i}}}2^{k_{i}+1}}=0
\end{eqnarray*}
Then $\lim_{i\rightarrow\infty}n_{k_i}^{1/2^{k_i}}=1$, therefore, $\liminf_{k\rightarrow\infty}n_{k}^{1/2^{k}}\leq 1$
Therefore, $\left\{n_{k}\right\}$ must be of the form:

$\left\{n_{k}\right\}$ is a sequence such that $$n_{k+1}>n_{k}^{2}-n_{k}+1$$ for all sufficiently large $k$, and such that for some $N\in\mathbb{N}$, $n_k\geq 2^k$ when $k\geq N$.

Let us prove the statement for this type of sequences, it follows from theorem A of this paper taking $b_{k}=1$ and $a_{k}=n_k$.
