If $(a_n)$ is increasing and $a_n^{1/c^n}\to\infty$, then $\sum\frac1{a_n}$ is irrational? I $\DeclareMathOperator\lcm{lcm}$am trying to generalise the result from this question: If $(a_n)$ is increasing and $\lim_{n \to \infty} a_n^{1/2^n} = \infty$, show that $\sum_{k=1}^{\infty} 1/{a_n}$ is irrational and I try to prove that

Conjecture (strong). If $(a_n)$ is an increasing natural sequence and $c>1$ such that $a_n^{1/c^n}\to\infty$, then $\sum\frac1{a_n}$ is irrational. (Note that for $c\geq2$ we are done already.)

Though I will also be happy if I only manage to prove that

Conjecture (weak). If $(a_n)$ is an increasing natural sequence such that $a_n^{1/c^n}\to\infty$ for all $1<c<2$, then $\sum\frac1{a_n}$ is irrational.

In particular I am wondering whether $\sum_{n=0}^\infty\frac1{2^{2^n}+1}$ is irrational. (just curiosity)

I'll summarize the proof in the linked question: It suffices to prove that
$$\liminf_{k\to\infty}\sum_{n=k+1}^\infty\frac{\lcm(a_1,\ldots,a_k)}{a_n}=0;$$
we'll prove that in fact
$$\liminf_{k\to\infty}\sum_{n=k+1}^\infty\frac{a_1\cdots a_k}{a_n}=0.$$
Let $a_n=b_n^{2^n}$ (so $b_n\to\infty$) and let $r$ be such that $b_n\leq b_r$ for all $n\leq r$. Then
$$\frac{a_1\cdots a_{r-1}}{a_{r+n}}=\frac{b_1^2\cdots b_{r-1}^{2^{r-1}}}{b_{r+n}^{2^{r+n}}}\leq\frac{b_r^{2+\cdots+2^{r-1}}}{b_{r+n}^{2^{r+n}}}\leq b_{r+n}^{2^n(2^r-2)-2^{r+n}}=b_{r+n}^{-2^{n+1}}.$$
Now choose $r$ such that $b_n\geq M>1$ for all $n\geq r$ and we get
$$\sum_{n=r}^\infty\frac{a_1\cdots a_{r-1}}{a_n}\leq\sum_{n=0}^\infty M^{-2^{n+1}}\leq\sum_{n=0}^\infty M^{-(n+1)}=\frac1M.$$
Because $M>1$ was arbitrary, we get
$$\liminf_{k\to\infty}\sum_{n=k+1}^\infty\frac{\lcm(a_1,\ldots,a_k)}{a_n}=0.$$

It's not hard to see that this strategy (proving $\liminf_{k\to\infty}\sum_{n=k+1}^\infty\frac{\lcm(a_1,\ldots,a_k)}{a_n}=0$) does not work for $c<2$. For example for $a_n=2^{2^n}+1$ we have $\lcm(a_0,\ldots,a_k)=a_1\cdots a_k=2^{2^{k+1}}-1\approx a_{k+1}^2$ so even controlling the size of the first term is hopeless. What else can we do?

Note: using the same strategy as in the linked question I was able to control the terms of $\sum_{n=k+1}^\infty\frac{a_1\cdots a_k}{a_n}$ (for suitable $k$) except the first $l-1$, where $l$ is such that $c^l\geq c^{l-1}+1$. I'm not mentioning the proof here because controlling the first $l-1$ terms seems hopeless anyway.
 A: 1.) The sum $\displaystyle \sum_{n=0}^\infty \frac{1}{2^{2^n}+1}$ is the sum of the reciprocals of the Fermat Numbers, and is proven irrational as a corollary of a beautiful theorem here. 
2.) It is unfortunate that even the weak conjecture is false, and a sequence given in the paper linked above can be proven to actually be a counterexample. The sequence is defined recursively by: $b_1 \doteq 2$, $b_{n+1} \doteq b_n^2-b_n+1$ for $n \geq 1$. We have from the paper that $\displaystyle \sum_{n=1}^\infty \frac{1}{b_n}=1 \in \mathbb{Q}$.  I claim that:
For every $1<c<2$, $\displaystyle \lim_{n \to \infty} b_n^{1/c^n}= +\infty$.
$\textit{Proof:}$  It is clear than $b_n \rightarrow \infty$. Then, since $1<c<2$, we may choose some $N \in \mathbb{N}$ such that for $n>N$, we have $b_n^2-b_n+1>b_n^c$. Now, note that for $n>N$:
$\frac{b_{n+1}^{1/c^{n+1}}}{b_n^{1/c^n}}= \frac{ ((b_n^2-b_n+1)^{1/c})^{1/c^n}}{b_n^{1/c^n}}> \frac{b_n^{1/c^n}}{b_n^{1/c^n}}=1$
So the sequence $S= (b_n^{1/c^n})_{n>N}$ increases monotonically. We then have either $S \rightarrow \ell \in \mathbb{R}$ or $S \rightarrow \infty$. Suppose, for a contradiction, that the former holds. Then $\ell$ must have the following property:
$\displaystyle \lim b_{n+1}^{1/c^{n+1}}= \ell= \displaystyle \lim (b_n^{1/c^n})^{2/c} \displaystyle \lim (1- \frac{1}{b_n} + \frac{1}{b_n^2})^{1/c^{n+1}}= \ell^{2/c}$
i.e. $\ell(1- \ell^{2/c-1})=0$, so $\ell=0$ or $\ell=1$ since $c \neq 2$. But the terms of $S$ are all larger than $1$ and increase monotonically, so neither of these options can hold. It follows that $S \rightarrow \infty$. $\Box$
