<s>[INST] Evaluating $\sum_{n=1}^\infty \frac{1}{(n^5)!} \approx 1$ and proving that is irrational. 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:
$\delta \approx \sum_{n=1}^{10} 1/(n^5)! \approx 1.000000000000000000000000000000000003800390754854743592594...$
 [/INST] 
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.)
</s>