If $(n_k)$ is strictly increasing and $\lim_{n \to \infty} n_k^{1/2^k} = \infty$ show that $\sum_{k=1}^{\infty} 1/n_k$ is irrational 
Prove that for a strictly increasing natural sequence $(n_k) $ satisfying $\lim_{n \to \infty} n_k^{1/2^k}=\infty$, $\sum_{k=1}^{\infty} 1/n_k$ is irrational.

This is another problem "problems in mathematical analysis" by Piotr and Witkowski. I've been trying to solve its problems for the last couple of days and I have thought about this one for a long time but nothing comes to my mind at all.
I can't even imagine how $\displaystyle \lim_{n \to \infty} n_k^{1/2^k}=\infty$ and $\displaystyle \sum_{k=1}^{\infty} 1/n_k$ could possibly be related to each other.
 A: Assume $\sum_{k=1}^{\infty} 1/n_k=\frac{p}{q}$ where $p,q$ are strictly positive integer,
Then,
$$
n_1n_2\cdots n_{k-1}\sum_{j=0}^{\infty}1/n_{k+j}\geq \frac{1}{q}
$$
for $k\geq 2$.
Now let $a_k=n_k^{\frac{1}{2^k}}$, by hypothesis $\lim_{n \to \infty} n_k^{1/2^k}=+\infty$
Therefore there exist $r_1$ such that $a_j\leq a_{r_1}$ for $j=1,2,\cdots,r_1-1$
Indeed, if $a_1\leq a_2$, $r_1=2$,
If this is not the case, we chooses the smallest integer $r_1>2$ such that $a_1\leq a_{r_1}$.
There are actually a whole infinite sequence $r_k$ checking the above property, that is to say such as,
$$
a_j\leq a_{r_k}, \quad j=1,2,\cdots,r_k-1
$$
To find $r_2$, the above procedure is applied to the following sequence $\{a_k\}_{k \geq r_1}$, and so forth.
We denote $r$ the smallest positive integer such that $a_{r+j}>q+1$ for $j\in \mathbb{N}$
and $a_j \leq a_r$ for $j=1,2,\cdots,r-1$.
Since, $n_r \leq n_{r+j}$  we have $a_r \leq a_{r+j}^{\large 2^j}$ 
$$
\frac{n_1n_2\cdots n_{r-1}}{n_{r+j}} \leq \frac{a_r^{\large 2+2^2+\cdots+2^{r-1}}}{n_{r+j}}\leq  \frac{a_{r+j}^{\large 2^{j}(2^r-2)}}{a_{r+j}^{\large 2^{r+j}}}=a_{r+j}^{\large -2^{j+1}}<(q+1)^{-(j+1)}
$$
Thus,
$$
n_1n_2\cdots n_{r-1}\sum_{j=0}^{\infty}1/n_{r+j}<\sum_{j=0}^{\infty}(q+1)^{-(j+1)}=\frac{1}{q}
$$
Contradiction $\square$

Addendum  : For the intuition behind the exercise, try this :
Let $\{n_k\}$ a sequence of strictly positive integer such that 
$$
\limsup_{k\rightarrow +\infty} \frac{n_k}{n_1n_2\cdots n_{k-1}}=+\infty, \quad \liminf_{k\rightarrow +\infty} \frac{n_k}{n_{k-1}}>1
$$
Show that 
$$
\sum_{i=0}^{\infty}\frac{1}{n_{i}}
$$
is irrational.
To prove it, assume $\sum_{k=1}^{\infty} 1/n_k=\frac{p}{q}$ and write
$$
\sum_{k=k_1}^{\infty} \frac{1}{n_k}=\frac{p}{q}-\sum_{k=1}^{k_1-1} \frac{1}{n_k}
$$


*

*What can you say about $\sum_{k=k_1}^{\infty} \frac{q n_1n_2\cdots n_{k-1}}{n_k}$ ?


Then the intuition behind my first inequality follow.
For the remainder of the proof, I will leave it to you (as planned).
