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?

  • $\begingroup$ The second serie is just a specialization of the first one, but I have no idea of how to deal with the first one. $\endgroup$ – Patrick Da Silva Nov 23 '11 at 1:19
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    $\begingroup$ @Kou How did this problem arise ? Please quote the source. $\endgroup$ – Sasha Nov 23 '11 at 2:27
  • $\begingroup$ This could be quite delicate: if you change $<$ to $\le$ in the displayed condition then the statement would no longer be true. Let $b_1=2$, $b_2=3$, $b_3=7$, $b_4=43$, and in general $b_{k+1} = b_k(b_k-1)+1$. Then $\sum b_k^{-1} = 1$ is rational, while $1/b_{k+1}+1/b_{k+2}+\cdots = 1/(b_k-1) - 1/b_k$ for each $k$. $\endgroup$ – Greg Martin Nov 23 '11 at 7:48

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.


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