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?
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.