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Let $(a_n)_{n=0}^{\infty}$ be a sequence of zeroes and ones such that $a_n=1$ for infinitely many $n$. Let $\displaystyle x:=\sum_{n=0}^{\infty} \frac{a_n}{n!} .$

Is $x$ irrational? I believe it is, but I don't know how to prove it. I'll appreciate any help.

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    $\begingroup$ I think this proof for the irrationality of $e$ should also work here, just change $\frac{1}{n!}$ to $\frac{a_n}{n!}$ $\endgroup$
    – leoli1
    Commented Feb 11, 2021 at 18:29
  • $\begingroup$ I see, thank you. $\endgroup$
    – jenda358
    Commented Feb 12, 2021 at 10:27
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    $\begingroup$ OK, next prove it is transcendental. $\endgroup$
    – GEdgar
    Commented Feb 12, 2021 at 12:04

1 Answer 1

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As was pointed out by leoli1, this problem can be easily solved by slightly modifying Fourier's proof of irrationality of Euler's number.

Case closed.

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