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This is the sum

$$\sum_{n=1}^{\infty}\frac{1}{{5^n}^!}$$

My first attempt was to assume that the series does converge to a rational number $a/b$. But the $n!$ bothered me and I failed in my proof.

How would you try to prove this series?

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Because in base $5$ that number will have a non-periodic expansion.

You can also prove it directly by using that the distance between non-zero digits increases. This means that you can for every $\epsilon>0$ find an (non-zero) integer $a$ such that $a\sum 5^{-n!}$ can be written as $N+\xi$ where $N$ is a natural number and $0<\xi<\epsilon$. If $k\sum 5^{-n!}$ is a natural number then so is $ak\sum 5^{-n!} = kN + k\xi$ where $0<\xi<k\epsilon<1$. Similar argument can be used to show that the number is transcendental.

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You can prove this with the approach often used with $e$ as well.

Thm: If $\xi\in\mathbb{Q}$ then there exists a $c>0$ depending only on $\xi$ s.t. when $p,q$ integers with $q>0$ and $\frac{p}{q}\neq\xi$, then $\left\lvert\xi -\frac{p}{q}\right\rvert\geq\frac{c}{q}$.

Proof: Choose $b,q>0$ then $\left\lvert\frac{a}{b} -\frac{p}{q}\right\rvert\geq\frac{1}{bq}$ (combine the fractions together to see this).

Let \begin{align*} \xi&=\sum_{k=1}^\infty 5^{-k!} \\ \xi_n&=\sum_{k=1}^n 5^{-k!} \end{align*}

The (reduced) denominator of $\xi_n$ is at most $5^{n!}$ and so let $p=5^{n!}\xi_n$ and $q=5^{n!}$ be our integers so that

$$\frac{c}{5^{n!}}\leq\left\lvert\xi -\xi_n\right\rvert=\sum_{k=n+1}^\infty 5^{-k!}\leq\sum_{k=(n+1)!}^\infty 5^{-k}=\frac{5}{4}\cdot 5^{-(n+1)!}=\frac{1}{4\cdot 5^{n}\cdot 5^{n!}}$$

and so we conclude that if $\xi\in\mathbb{Q}$,$$\frac{c}{5^{n!}}\leq \frac{1}{4\cdot 5^{n}\cdot 5^{n!}}$$

this cannot be true for all $n$ no matter the $c$, and so $\xi\notin\mathbb{Q}$

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  • $\begingroup$ Thanks, I started to read more about it, but what theorem is that? $\endgroup$ – O. Ewen Jul 28 '17 at 4:10

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