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

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).