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I'm trying to prove that

$$ \sum_{k=1}^{\infty} 7^{-k!} $$

is irrational but I'm so lost. Any tips for where to begin, thanks in advance.

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marked as duplicate by Hans Lundmark, Dando18, Daniel W. Farlow, Siong Thye Goh, Claude Leibovici Aug 3 '17 at 6:45

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    $\begingroup$ if a number is rational, then in any base, the "digits" of it will either terminate or ultimately become periodic. $\endgroup$ – achille hui Jul 27 '17 at 18:04
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    $\begingroup$ That'a a Liouville number, so it is not only irrational, it is transcendental. The usual method of proving any Liouville number irrational is by contradiction. $\endgroup$ – NickD Jul 27 '17 at 18:19
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$$\sum_{k=1}^{\infty} 7^{-k!} = \frac{1}{7} + \frac{1}{7^{2!}} + \frac{1}{7^{3!}} + \dots$$ has a base 7 representation of $(0.11000100.....1000000.............1000000000......)_7$ where there is a $1$ at every $n!$th place from the radix point, and $0$s at the rest of the places.

A real number is rational if and only if its positional representation either terminates or repeats in any base.

This series converges to a number whose base 7 representation does not repeat (clearly), or terminate. Therefore, it is irrational.

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