The decimal expansion is A331373 in the OEIS. Weisenberg has observed that this sum can also be written as \[\sum_{k\geq 1}\sum_{n\geq 2}\frac{1}{(n!)^k}.\]

Erdős [Er48] proved that $\sum_n \frac{d(n)}{2^n}$ is irrational, where $d(n)$ is the divisor function.

Pratt [Pr24] has proved this is irrational, conditional on a uniform version of the prime $k$-tuples conjecture.

Tao has observed that this is a special case of [257], since \[\sum_{n\geq 2}\frac{\omega(n)}{2^n}=\sum_p \frac{1}{2^p-1}.\]

Erdős [Er75c] proved the answer is yes under the stronger condition that $\limsup n_k/k^t=\infty$ for all $t\geq 1$.

The decimal expansion of this sum is A256936 on the OEIS.

The answer is yes, as shown by Nesterenko [Ne96].

The decimal expansion of this sum is A098990 on the OEIS.

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].

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 Erdős [Er48] proved is irrational.

The case when $A$ is the set of primes is [69].

Erdős and Straus [ErSt71] have proved this is true if $a_n$ is monotone, i.e. $a_{n-1}\leq a_n$ for all $n$. Erdős [Er48] proved that $\sum_n \frac{d(n)}{t^n}$ is irrational for any integer $t\geq 2$.

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.

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 $a_n^{1/n}\to\infty$.

This was essentially solved by Hančl [Ha91], who proved that such a sequence needs to satisfy \[\limsup_{n\to \infty} \frac{\log_2\log_2 a_n}{n} \geq 1.\] More generally, if $a_n\ll 2^{2^{n-F(n)}}$ with $F(n)<n$ and $\sum 2^{-F(n)}<\infty$ then $a_n$ cannot be an irrationality sequence.

One possible definition of an 'irrationality sequence' (see also [262] and [264]). A folklore result states that $\sum \frac{1}{a_n}$ is irrational whenever $\lim a_n^{1/2^n}=\infty$.

Kovač and Tao [KoTa24] have proved that any strictly increasing sequence such that $\sum \frac{1}{a_n}$ converges and $\lim a_{n+1}/a_n^2=0$ is not such an irrationality sequence. On the other hand, if \[\liminf \frac{a_{n+1}}{a_n^{2+\epsilon}}>0\] for some $\epsilon>0$ then the above folklore result implies that $a_n$ is such an irrationality sequence.

A possible definition of an 'irrationality sequence' (see also [262] and [263]). One example is $a_n=2^{2^n}$. In [ErGr80] they also ask whether such a sequence can have polynomial growth, but Erdős later retracted this in [Er88c], claiming 'It is not hard to show that it cannot increase slower than exponentially'.

Kovač and Tao [KoTa24c] have proved that $2^n$ is not such an irrationality sequence. More generally, they prove that any strictly increasing sequence of positive integers such that $\sum\frac{1}{a_n}$ converges and \[\liminf \left(a_n^2\sum_{k>n}\frac{1}{a_k^2}\right) >0 \] is not such an irrationality sequence. In particular, any strictly increasing sequence with $\limsup a_{n+1}/a_n <\infty$ is not such an irrationality sequence.

On the other hand, Kovač and Tao do prove that for any function $F$ with $\lim F(n+1)/F(n)=\infty$ there exists such an irrationality sequence with $a_n\sim F(n)$.

Cantor observed that $a_n=\binom{n}{2}$ is such a sequence. If we replace $-1$ by a different constant then higher degree polynomials can be used - for example if we consider $\sum_{n\geq 2}\frac{1}{a_n}$ and $\sum_{n\geq 2}\frac{1}{a_n-12}$ then $a_n=n^3+6n^2+5n$ is an example of both series being rational.

Erdős believed that $a_n^{1/n}\to \infty$ is possible, but $a_n^{1/2^n}\to 1$ is necessary.

This has been almost completely solved by Kovač and Tao [KoTa24], who prove that such a sequence can grow doubly exponentially. More precisely, there exists such a sequence such that $a_n^{1/\beta^n}\to \infty$ for some $\beta >1$.

It remains open whether one can achieve \[\limsup a_n^{1/2^n}>1.\] A folklore result states that $\sum \frac{1}{a_n}$ is irrational whenever $\lim a_n^{1/2^n}=\infty$, and hence such a sequence cannot grow faster than doubly exponentially - the remaining question is the precise exponent possible.

This conjecture is due to Stolarsky.

A negative answer was proved by Kovač and Tao [KoTa24], who proved even more: there exists a strictly increasing sequence of positive integers $a_n$ such that \[\sum \frac{1}{a_n+t}\] converges to a rational number for every $t\in \mathbb{Q}$ (with $t\neq -a_n$ for all $n$).

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 sum $\sum \frac{1}{F_n}$ itself was proved to be irrational by André-Jeannin [An89].

If $P$ is infinite this sum is always 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.

Crmarić and Kovač [CrKo25] have shown that the answer to this question is no in a strong sense: for any $\alpha \in (0,\infty)$ there exists a function $f:\mathbb{N}\to\mathbb{N}$ such that $f(n)\to \infty$ as $n\to\infty$ and \[\sum_{n\geq 1} \frac{1}{(n+1)\cdots (n+f(n))}=\alpha.\] It is still possible that this sum is always irrational if $f$ is assumed to be non-decreasing; Crmarić and Kovač show that the set of the possible values of such a sum has Lebesgue measure zero.