TITLE: Irrationality of Two Series
QUESTION [11 upvotes]: Show that if the integers $1<b_1<b_2<\cdots$ increase so rapidly that$$\frac{1}{b_{k+1}}+\frac{1}{b_{k+2}}+\cdots<\frac{1}{b_{k}-1}-\frac{1}{b_{k}},\quad k\geq 1,$$
  then the number $\sum b_k^{-1}$ is irrational. Prove that $\sum_{0}^{\infty}(2^{3^k}+1)^{-1}$ is irrational.

I don't know how to think about them. Could anyone prove them?

REPLY [5 votes]: See if references in this post by Ragib Zaman are useful.
Specifically, the following theorem by Erdős, says that for an increasing sequence $a_k$ of positive integers, in this case $a_k = 2^{3^k}+1$, such that $\underset{n\to \infty}{\lim \sup}\; a_n^{\frac1{2^n}} = \infty$ and $a_n > n^{1+\epsilon}$ for every $\epsilon > 0$ and $n>n_0(\epsilon)$, then the sum $\sum\limits_{n=1}^\infty \frac1{a_n}$ is an irrational number.
It is easy to check that these conditions are met:
$$
  \underset{n\to \infty}{\lim \sup} \left( 2^{3^n}+1 \right)^{\frac1{2^n}} = 
   \lim_{n \to \infty} 2^{(3/2)^n} = \infty
$$
Also clearly $a_n$ grows faster than $n^{1+\epsilon}$ for any $\epsilon > 0$, thus it follows that $\sum\limits_{n=1}^\infty \frac1{2^{3^n}+1}$ is irrational.