Prove the series $ \sum_{n=1}^\infty \frac{1}{(n!)^2}$ converges to an irrational number How can one prove that the series $\displaystyle \sum\limits_{n=1}^\infty \frac{1}{(n!)^2}$ converges to an irrational number? There's no need to use Taylor expansion, integrals or any advanced/professional techniques. It can be proved using only basic techniques.
My first attempt was to assume that the series does converge to a rational number $ \frac{p}{q}$ and then it follows by definition that for every positive epsilon there exists an integer k such that $ {\sum_{n=1}^{k} \frac{1}{(n!)^2} -\frac{p}{q}}<=\epsilon$
I tried to get a contradiction but I failed.
 A: Here is a tiny proof that $e$ is irrational. Your question can be solved the same way.
A: Hint: if $x=\frac pq$, what can you say about $x(q!)^2$?
Another Hint: what can you say about $(q!)^2\sum\limits_{n=1}^q\frac1{(n!)^2}$? what about $(q!)^2\sum\limits_{n=q+1}^\infty\frac1{(n!)^2}$?
A: Firstly I'd advise you to go over the proof in "Walter Rudin: Principles of Mathematical Analysis", chapter on Sequences and series, where he proves that e is irrational. Then Use the same(similar) steps and show the following:
If $$\sum_{k=1}^\infty \frac{1}{(k!)^2} = L$$ and $$\sum_{k=1}^n \frac{1}{(k!)^2} = L_n$$
$$0 < L - L_n < \frac{1}{(n!)^2 n(n+2)}$$
Follow similar steps to get your result.
A: Proposition. The following series converges 
$$\sum_{n=1}^{\infty}\frac{1}{(n!)^2},$$
to a number which we call $S$. The number $S$ lies in the interval 
$$1.2795<S<1.2796$$ and is irrational.
Proof. To examine whether the series converges, let us examine if the tail of the series tends to zero. For any integers $0<N<M,$
one has
$$\sum_{n=N+1}^{M}\frac{1}{n!^2}=\frac{1}{(N+1)!^2}+
\frac{1}{(N+2)!^2}+\frac{1}{(N+3)!^2}+\cdots$$
$$=\frac{1}{(N+1)!^2}\Big[1+\frac{1}{(N+2)^2}+\frac{1}{(N+2)^2(N+3)^2}+\cdots\Big],
$$
where the summation in each line stops at the term corresponding to $M,$
but which we do not write down for presentation purposes. The quantity inside the brackets is smaller than
$$1+\frac{1}{(N+2)^2}+\frac{1}{(N+2)^4}+\cdots
$$ which is a geometric sum with ratio $$x:=\frac{1}{(N+2)^2}.$$
Since $0<x<1,$ the geometric series converges and we can extend the 
summation to infinity to get that the brackets quantity is smaller than
$$\frac{1}{1-\frac{1}{(N+2)^2}}
\leq 2.$$
We have proved that for any $0<N<M,$ we have 
$$0<\sum_{n=N+1}^{M}\frac{1}{(n!)^2}< \frac{2}{(N+1)!^2} \to 0,$$
and therefore our series converges.
To show that the number defined by the series lies in $[1.2795,1.2796]$
we take $N=4$ in the following inequality, which we just proved:
$$\sum_{n=1}^{N}\frac{1}{(n!)^2} < S <
\sum_{n=1}^{N}\frac{1}{(n!)^2}
+
\frac{2}{(N+1)!^2}.
$$
We conclude by showing the irrationality of S. Suppose to the contrary that $S$ is rational, i.e. there exist integers $a,b$ such that $S=\frac{a}{b}.$ Since   $S>0,$ we get that $a,b$ have the same sign. We can assume that they are both positive, as we can multiply numerator and denominator of the fraction by $-1$
if they are not. Note that since $S>1,$ we have that $a>b\geq 1.$
We next get for $N=b,$
that
$$
0<S-\sum_{n=1}^{b}\frac{1}{(n!)^2}<\frac{2}{(b+1)!^2},
$$ so that by using $S=a/b$ and multiplying by $b!^2,$
we get
$$
0<a (b-1)!b!-\sum_{n=1}^{b}\frac{(b!)^2}{(n!)^2}<\frac{2}{(b+1)}\frac{1}{(b+1)}.
$$
We will now arrive at a contradiction by proving that the number in the middle is an integer and the number on the right is $<1.$
The first assertion is proved by observing that whenever $n\leq b,$
then $n!$ divides $b!$ and the second assertion is proved by noticing that
$2 \leq b+1. \ \ \ \ \ \square $
EDIT: One can prove a similar statement using the same method, namely that whenever $a_n$ is a strictly increasing sequence of natural numbers, then the series $$\sum_{n=1}^{\infty}\frac{1}{a_n!}$$
converges to an irrational number. The case $a_n=n$ corresponds to $e.$ In this way one can give explicit constructions of uncountably many irrationals. To prove irrationality, if $S=a/b$ as before, let $N$ be defined as the least positive integer with the property that $$a_N \geq b.$$ Then one gets 
$$0<a\frac{a_N!}{b}-\sum_{n=1}^{N}\frac{a_N!}{a_n!}\leq
a_N!\sum_{k=1+a_N}^{\infty}\frac{1}{k!}
<a_N!\frac{2}{(1+a_N)!}=\frac{2}{1+a_N}\leq 1.$$
A: Write it as $p/q$ with $p\,q$ coprime integers, $q\le2$, so $q!(q-1)!p=\sum_{n\ge1}\frac{q!^2}{n!^2}$. Hence $\sum_{n\ge q+1}\frac{q!^2}{n!^2}$ is an integer. But it can't be, as it's a positive number $\le\sum_{k\ge1}\frac{1}{q^{2k}}=\frac{1}{q^2-1}$.
