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Is \[\sum_{n\geq 2}\frac{1}{n!-1}\] irrational?
Is \[\sum_{n\geq 2}\frac{\omega(n)}{2^n}\] irrational? (Here $\omega(n)$ counts the number of distinct prime divisors of $n$.)
Erdős [Er48] proved that $\sum_n \frac{d(n)}{2^n}$ is irrational, where $d(n)$ is the divisor function.
Let $n_1<n_2<\cdots$ be a sequence of integers such that \[\limsup \frac{n_k}{k}=\infty.\] Is \[\sum_{k=1}^\infty \frac{1}{2^{n_k}}\] transcendental?
Erdős [Er75c] proved the answer is yes under the stronger condition that $\limsup n_k/k^t=\infty$ for all $t\geq 1$.
Is \[\sum_n \frac{\phi(n)}{2^n}\] irrational? Here $\phi$ is the Euler totient function.
Is \[\sum \frac{\sigma(n)}{2^n}\] irrational? (Here $\sigma(n)$ is the sum of divisors function.)
The answer is yes, as shown by Nesterenko [Ne96].
Is \[\sum \frac{p_n}{2^n}\] irrational? (Here $p_n$ is the $n$th prime.)
Let $k\geq 1$ and $\sigma_k(n)=\sum_{d\mid n}d^k$. Is \[\sum \frac{\sigma_k(n)}{n!}\] irrational?
This is known now for $1\leq k\leq 4$. The cases $k=1,2$ are reasonably straightforward, as observed by Erdős [Er52]. The case $k=3$ was proved independently by Schlage-Puchta [ScPu06] and Friedlander, Luca, and Stoiciu [FLC07]. The case $k=4$ was proved by Pratt [Pr22].
Let $A\subseteq \mathbb{N}$ be an infinite set. Is \[\sum_{n\in A}\frac{1}{2^n-1}\] irrational?
If $A=\mathbb{N}$ then this series is $\sum_{n}\frac{d(n)}{2^n}$, where $d(n)$ is the number of divisors of $n$, which is known to be irrational.
Let $a_n\to \infty$. Is \[\sum_{n} \frac{d(n)}{a_1\cdots a_n}\] irrational, where $d(n)$ is the number of divisors of $n$?
Erdős and Straus [ErSt71] have proved this is true if $a_n$ is monotone, i.e. $a_{n-1}\leq a_n$ for all $n$.
Is the sum \[\sum_{n} \mu(n)^2\frac{n}{2^n}\] irrational?
Additional thanks to: Boris Alexeev and Dustin Mixon
Let $a_n$ be a sequence such that $a_n/n\to \infty$. Is the sum \[\sum_n \frac{a_n}{2^{a_n}}\] irrational?
This is true under either of the stronger assumptions that
  • $a_{n+1}-a_n\to \infty$ or
  • $a_n \gg n\sqrt{\log n\log\log n}$.
Erdős and Graham speculate that the condition $\limsup a_{n+1}-a_n=\infty$ is not sufficient, but know of no example.
Suppose $a_1<a_2<\cdots$ is a sequence of integers such that for all integer sequences $t_n$ with $t_n\geq 1$ the sum \[\sum_{n=1}^\infty \frac{1}{t_na_n}\] is irrational. How slowly can $a_n$ grow?
One possible definition of an 'irrationality sequence' (see also [263] and [264]). An example of such a sequence is $a_n=2^{2^n}$, while a non-example is $a_n=n!$. It is known that if $a_n$ is such a sequence then $a_n^{1/n}\to\infty$.
Let $a_n$ be a sequence of integers such that for every sequence of integers $b_n$ with $b_n/a_n\to 1$ the sum \[\sum\frac{1}{b_n}\] is irrational. Is $a_n=2^{2^n}$ such a sequence? Must such a sequence satisfy $a_n^{1/n}\to \infty$?
One possible definition of an 'irrationality sequence' (see also [262] and [264]).
Let $a_n$ be a sequence of integers such that, for every bounded sequence $b_n$, the sum \[\sum \frac{1}{a_n+b_n}\] is irrational. Are $a_n=2^n$ or $a_n=n!$ examples of such a sequence? Is there such a sequence with $a_n<n^k$?
A possible definition of an 'irrationality sequence' (see also [262] and [263]). One example is $a_n=2^{2^n}$.
How fast can $a_n\to \infty$ grow if \[\sum\frac{1}{a_n}\quad\textrm{and}\quad\sum\frac{1}{a_n+1}\] are both rational?
Cantor observed that $a_n=\binom{n}{2}$ is such a sequence. If we replace $+1$ by a larger constant then higher degree polynomials can be used - for example if we consider $\sum\frac{1}{a_n}$ and $\sum\frac{1}{a_n+8}$ then $a_n=n^3+6n^2+5n$ is an example.
Let $a_n$ be an infinite sequence of integers. There exists some integer $t\geq 1$ such that \[\sum \frac{1}{a_n+t}\] is irrational.
This conjecture is due to Stolarsky.
Let $F_1=F_2=1$ and $F_{n+1}=F_n+F_{n-1}$ be the Fibonacci sequence. Let $n_1<n_2<\cdots $ be an infinite sequence with $n_{k+1}/n_k \geq c>1$. Must \[\sum_k\frac{1}{F_{n_k}}\] be irrational?
It may be sufficient to have $n_k/k\to \infty$. Good [Go74] and Bicknell and Hoggatt [BiHo76] have shown that $\sum \frac{1}{F_{2^n}}$ is irrational. The nature of $\sum \frac{1}{F_n}$ itself is unknown.
Let $P$ be a finite set of primes with $\lvert P\rvert \geq 2$ and let $\{a_1<a_2<\cdots\}=\{ n\in \mathbb{N} : \textrm{if }p\mid n\textrm{ then }p\in P\}$. Is the sum \[\sum_{n=1}^\infty \frac{1}{[a_1,\ldots,a_n]},\] where $[a_1,\ldots,a_n]$ is the lowest common multiple of $a_1,\ldots,a_n$, rational or irrational?
If $P$ is infinite this sum is always irrational.
Let $f(n)\to \infty$ as $n\to \infty$. Is it true that \[\sum_n \frac{1}{(n+1)\cdots (n+f(n))}\] is irrational?
Erdős and Graham write 'the answer is almost surely in the affirmative if $f(n)$ is assumed to be nondecreasing'. Even the case $f(n)=n$ is unknown, although Hansen [Ha75] has shown that \[\sum_n \frac{1}{\binom{2n}{n}}=\sum_n \frac{n!}{(n+1)\cdots (n+n)}=\frac{1}{3}+\frac{2\pi}{3^{5/2}}\] is transcendental.