Evaluating $\sum_{n=1}^\infty \frac{1}{(n^5)!} \approx 1$ and proving that is irrational.

Question

Define $\delta = \sum_{n=1}^\infty \frac{1}{(n^5)!}$. Wolfram says it converges by the ratio test. Trying to prove that $\delta$ is irrational, begin defining $S_n$ as:

\begin{align} S_n = (n^5)!\delta \: - (n^5)!\sum_{k=1}^n\frac{1}{(k^5)!} \end{align}

Where $(n^5)!\sum_{k=1}^n\frac{1}{(k^5)!}$ is an integer. Write $\delta = 1/(1^5)!+1/(2^5)! + 1/(3^5)!+...+1/(n^5)!+1/(n+1)^5!+...$, so

\begin{align} S_n &= (n^5)!\delta \: - (n^5)!\sum_{k=0}^n\frac{1}{(k^5)!}\\ &=(n^5)!\left(\delta - \sum_{k=0}^n\frac{1}{(k^5)!} \right)\\ &=(n^5)! \left( \frac{1}{(1^5)!}+\frac{1}{(2^5)!}+...+\frac{1}{(n^5)!}+ \frac{1}{(n+1)^5!}+... - \sum_{k=1}^n\frac{1}{(k^5)!} \right) \end{align}

expanding the sum on the right it's possible to cancel a few terms:

\begin{align} S_n &= (n^5)! \left( \frac{1}{(1^5)!}+\frac{1}{(2^5)!}+...+\frac{1}{(n^5)!}+ \frac{1}{(n+1)^5!}+... - \frac{1}{(1^5)!} - \frac{1}{(2^5)!}-...-\frac{1}{(n^5)!} \right)\\ &=(n^5)! \left( \frac{1}{(n+1)^5!}+\frac{1}{(n+2)^5!}+... \right)\\ &= \frac{(n^5)!}{(n+1)^5!}+\frac{(n^5)!}{(n+2)^5!}+... \\ \end{align}

From this post we have that

\begin{align} S_n = \frac{(n^5)!}{(n+1)^5!}+\frac{(n^5)!}{(n+2)^5!}+\cdots < \frac{1}{n^5+1} \end{align}

So we have that $0<S_n<\frac{1}{n^5+1}$. Assume that $\delta = p/q$ then: \begin{align} 0< (n^5)!p \: - q(n^5)!\sum_{k=1}^n\frac{1}{(k^5)!} < \frac{q}{n^5+1} \end{align}

So, for a large $n$, we found a integer between $0$ and $1$, meaning $\delta$ is irrational

Wolfram says that $\delta=1$, but then the proof above is wrong.

How to find $\delta$ analytically? Numerically, also by wolf:

$\delta \approx \sum_{n=1}^{10} 1/(n^5)! \approx 1.000000000000000000000000000000000003800390754854743592594...$

Answer 1

A Liouville number is a real number x with the property that, for every positive integer n, there exist integers p and q with q > 1 and such that $$ 0<\left|x-{\frac {p}{q}}\right|<{\frac {1}{q^{n}}}. $$ A Liouville number can thus be approximated "quite closely" by a sequence of rational numbers. In 1844, Joseph Liouville showed that all Liouville numbers are transcendental, thus establishing the existence of transcendental numbers for the first time.

( https://en.wikipedia.org/wiki/Liouville_number)

Your number is clearly a Liouville number.

(I haven't checked your proof.)