We should prove, that for all sequences $ \{a_i\}, a_i\in N$, such that $ 1<a_1<a_2<...$ \[ \sum_{a_i}^{}{\frac{2^{a_i}}{a_{i}!}}\] will be irrational number. (*) Let for all sequences it will be rational number. So: \[ \sum_{a_i}^{}{\frac{2^{a_i}}{a_{i}!}}=\frac{p}{q}\] Let sequence $ \{b_i\}, b_i\in N$, such that $ 1<b_1<b_2<...$ and $ b_i \ne a_j$. Using formila \[ e^x=\sum_{i=0}^{\infty}{\frac{x^n}{n!}}\] we will have: \[ e^2-1=\sum_{a_i}^{}{\frac{2^{a_i}}{a_{i}!}}+\sum_{b_i}^{}{\frac{2^{b_i}}{b_{i}!}}\] But \[ \sum_{a_i}^{}{\frac{2^{a_i}}{a_{i}!}}=\frac{p}{q}\] So, we have: \[ \sum_{b_i}^{}{\frac{2^{b_i}}{b_{i}!}}=e^2-1-\frac{p}{q}\] which is irrational, because $ e^2$ is irrational. This is contradiction with (*).

Unfortunately, this does not prove the thesis. The negation of the proposition "for all sequences $\{a_i\}$, $a_i \in \mathbb{N}$, with $1 < a_1 < a_2 < \cdots$, $\sum_{i=1}^{\infty} \frac {2^{a_i}} {a_i!}$ is irrational" is not (*) "for all sequences $\{a_i\}$, $a_i \in \mathbb{N}$, with $1 < a_1 < a_2 < \cdots$, $\sum_{i=1}^{\infty} \frac {2^{a_i}} {a_i!}$ is rational", so the above only proves that if, for such a sequence $\{a_i\}$, its associated series has a rational limit, then for the complementary sequence $\mathbb{N}^* \setminus \{a_i\}$, its associated series has an irrational limit.

I'm confused. I see, suppose its not irrational. Then it is rational. Bla bla. Contradiction. Hence its irrational. :S

Try to do this one: Let $S_{n} = \sum^{n} \frac{2^{a_{i}}}{a_{i}!}$. Then of course \[ S_{n+1} = \frac{(a_{1}!a_{2}!...a_{n}! \cdot A_{n})}{a_{n+1}!} \] where $A_{n} \in \mathbb{N}$ It's enough to prove that, there exist sequance $p_{n}, q_{n} \in \mathbb{N}$ such that $|S_{\infty}-p_{n}/q_{n}| = o(1/q_{n})$ But on the other hand \[ |S_{\infty} - S_{n}| = |S_{\infty} - \frac{(a_{1}!a_{2}!...a_{n-1}! (A_{n-1}))}{a_{n}!}| = \sum_{j=n+1}^{\infty}\frac{2^{a_{j}}}{a_{j}!} \] Try to prove, that \[ \lim_{n \to \infty} \frac{a_{n}!}{a_{n-1}!}\sum_{j=n+1}^{\infty}\frac{2^{a_{j}}}{a_{j}!} \to 0 \] Try to use Stirlnig formula. P.S. I'm not sure is it true or not, but it's one of the possible ways approach to such problem.

I will borrow a quill from Liouville's quiver. For a given sequence $0\leq a_1 < a_2 < \cdots$ of non-negative integers, let me denote by $\displaystyle E_n = \sum_{a_k \leq n} \dfrac {2^{a_k}} {a_k!}$. Since the sequence $(E_n)_{n\geq 1}$ is non-decreasing, and since $\displaystyle E_n < \sum_{k=0}^{\infty} \dfrac {2^k} {k!} = \textrm{e}^2$, it follows it is convergent to its supremum $\displaystyle E = \lim_{n\to \infty} \sum_{a_k \leq n} \dfrac {2^{a_k}} {a_k!}$. Assume $E = \dfrac {p} {q}$, with $p,q \in \mathbb{N}^*$, $\gcd(p,q) = 1$. Let us denote by $v_2(N)$ the exponent of $2$ in the canonical factorization of $N$, and take $\alpha = v_2(q)$. It is immediate from Legendre's formula that $v_2(n!) \leq n-1$, with equality if and only if $n=2^m$ for some positive integer $m$. Take then a large enough $m$ (thus a large enough $n$), so that $q \mid \dfrac {n!} {2^{n-1}} 2^{\alpha}$, hence $\dfrac {n!} {2^{n-1}} 2^{\alpha} E = \dfrac {n!} {2^{n-1}} 2^{\alpha}\dfrac {p} {q} \in \mathbb{N}^*$. On the other hand, it is clear that $\dfrac {n!} {2^{n-1}} 2^{\alpha} \dfrac {2^k} {k!} \in \mathbb{N}^*$ for all $0\leq k\leq n$, hence $\dfrac {n!} {2^{n-1}} 2^{\alpha} E_n \in \mathbb{N}^*$. Now, $\displaystyle 0< E-E_n \leq \sum_{k=n+1}^{\infty} \dfrac {2^k} {k!} < \sum_{k=n+1}^{\infty} \dfrac {2^k} {n!(n+1)^{k-n}}$, therefore $\displaystyle 0< \dfrac {n!} {2^{n-1}} 2^{\alpha} (E-E_n) <$ $ 2^{\alpha+1} \sum_{k=n+1}^{\infty} \dfrac {2^{k-n}} {(n+1)^{k-n}} =$ $ 2^{\alpha+1} \sum_{k=n+1}^{\infty} \dfrac {1} {((n+1)/2)^{k-n}} =$ $ 2^{\alpha + 1} \dfrac {2} {n+1} \cdot \dfrac {1} {1 - 2/(n+1)} =$ $ \dfrac {2^{\alpha+2}} {n-1} < 1$ for large enough $m$ (and so large enough $n$). On the other hand, according with the above, $\displaystyle \dfrac {n!} {2^{n-1}} 2^{\alpha} (E-E_n)$ should be an integer, thus establishing a contradiction. Re the above post: I do not think that asymptotic formulae (like Stirling's) can decide on the rationality of a limit expression.

mavropnevma wrote: Re the above post: I do not think that asymptotic formulae (like Stirling's) can decide on the rationality of a limit expression. I don't understand well why you think so. In my case, you just need to prove that the limit \[ \lim_{n\to\infty}\frac{a_{n}!}{a_{n-1}!}\sum_{j=n+1}^{\infty}\frac{2^{a_{j}}}{a_{j}!} \] is equal to zero. And one of the best way to show that is, to use stirling formula.

To start with, your claim Let $S_{n} = \sum^{n} \frac{2^{a_{i}}}{a_{i}!}$. Then of course \[ S_{n+1} = \frac{(a_{1}!a_{2}!...a_{n}! \cdot A_{n})}{a_{n+1}!} \] where $A_{n} \in \mathbb{N}$. is false. Just take $a_1=3, a_2=4,a_3=5$, and $n=2$, so $S_{2+1} = S_3 = \frac {2^3} {3!} + \frac {2^4} {4!} + \frac {2^5} {5!} = \frac {2^3\cdot 34} {5!}$, not of the form $\frac {3!\cdot 4!\cdot A_2} {5!}$.

yes, you are right, as always I made mistake in my calculations . Therefore Let me try to explain in more approximately language, since the sence was not in the exact calculation, it was just in approximations, but anyway my first approximation was not correct. since $ a_{j}!/a_{i}! = 2^{a_{j}-a_{i}-100}b_{i,j}$ for $j>i$, where $b_{i,j} \in \mathbb{N}$ hence $S_{n+1} =\frac{2^{a_{n+1}-100}b_{n+1} }{a_{n+1}!}$ where $b_{n+1} \in \mathbb{N}$ Then, if you again repeat the same things, it's enough to prove such limit \[ \lim_{n \to \infty} \frac{a_{n}!}{2^{a_{n}}}\sum_{j=1}^{\infty}\frac{2^{a_{n+j}}}{a_{n+j}!}=0 \] Which again seems to be true by stirling formula .

It's enough for what? You propose an estimate of a limit, without stating what is the use you propose to make of it. Being a little more exact, and clearer in your purposes will help everybody.

mavropnevma wrote: It's enough for what? You propose an estimate of a limit, without stating what is the use you propose to make of it. Being a little more exact, and clearer in your purposes will help everybody. For the irrationality of $\sum_{j=1}^{\infty} 2^{a_{j}}/a_{j}!$, it's enough to prove that the limit \[ \lim_{n\to\infty}\frac{a_{n}!}{2^{a_{n}}}\sum_{j=1}^{\infty}\frac{2^{a_{n+j}}}{a_{n+j}!} \] is equal to zero.

That is a statement, not a proof. If you will elaborate on it, you will see that it amounts to the proof I gave in full detail in a previous post.

Yes but your post was after my, therefore I was not read it. What you have done, you just reduced the problem with the same limit what I have written (because we are using one and the same things to proving irrationality). Yes I agree that you have written it more detailed proof (because Only now I read it). I just was posting some hints (with one error calculation ) and that is all, but anyway the idea was right. As you see (and have done) my limit is equal to zero, since \[ \frac{a_{n}!}{2^{a_{n}}}\sum_{j=1}^{\infty}\frac{2^{a_{j}}}{a_{j}!} < \frac{a_{n}!}{2^{a_{n}}}\sum_{j=a_{n+1}}^{\infty}\frac{2^{j}}{j!} < x_{n}:=\frac{(a_{n+1}-1)!}{2^{a_{n+1}-1}}\sum_{j=a_{n+1}}^{\infty}\frac{2^{j}}{j!} \to 0 \] since \[ A_{n}=\frac{n!}{2^{n}}\sum_{j=1}^{\infty}\frac{2^{j+n}}{(j+n)!} \to 0 \] for example by stirling formula (this is about what I was talking on each my posts and was answering on your first question), or for example by the simplest estimates what already have you written like this one \[ \frac{1}{(n+j)!} < \frac{1}{(n+1)!j!} \] And the sequence $x_{n}$ is the subsequence of $A_{n}$. But anyway, I remaind you that I just was answering on the your first question Extremal wrote: mavropnevma wrote: Re the above post: I do not think that asymptotic formulae (like Stirling's) can decide on the rationality of a limit expression. I don't understand well why you think so. hence this sentences mavropnevma wrote: If you will elaborate on it, you will see that it amounts to the proof I gave in full detail in a previous post. is more adressed to you not to me (because your solution was written after my post) (with one remarks: If you will elaborate on it, you will see that your detailed proof, it is detailed explanation of my hints I gave in my first post Extremal wrote: It's enough to prove that, there exist sequance $p_{n}, q_{n} \in \mathbb{N}$ such that $|S_{\infty}-p_{n}/q_{n}| = o(1/q_{n})$ and the proof of the limit Extremal wrote: it's enough to prove such limit \[ \lim_{n \to \infty} \frac{a_{n}!}{2^{a_{n}}}\sum_{j=1}^{\infty}\frac{2^{a_{n+j}}}{a_{n+j}!}=0 \] Which again seems to be true by stirling formula . .) )

You missed the most important point that for infinitely many $a$, the number $\frac{a!}{b!2^{a-b}}$ is an integer for all $1\le b<a$. Your tail estimate is the same as mavropnevma's but he proves that the beginning segment is an integer and you don't. Anyway, this discussion is sort of boring. How about the same problem with $3$ instead of $2$?

Well, Aigner and Ziegler, in their Proofs from the Book, claim that proving that $\textrm{e}^3$ is irrational (certainly a most particular case of what you asked, fedja) requires heavier machinery (bit of calculus), using the function $\frac {x^n(1-x)^n} {n!}$ with various differentiation and integration tricks. Will that idea hold for the "holey" sequence $a_i$ ?

I agree with mavropnevma and it seems to be open problem in the case 3. and not only in this case, also for all $m>2$