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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.

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Have you tried something? If so please post it in the body of the question. Also advanced/professional tools?? – Gautam Shenoy Apr 20 '13 at 9:12
You can prove your series converges to an irrational number using the same approach in this proof that e is irrational. – achille hui Apr 20 '13 at 9:38

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 $(q!)^2\sum\limits_{n=q+1}^\infty\frac1{(n!)^2}$?

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You probably mean $x(q!)^2$. – Ivan Loh Apr 20 '13 at 9:20
@IvanLoh: indeed – robjohn Apr 20 '13 at 9:34

Here is a tiny proof that $e$ is irrational. Your question can be solved the same way.

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Not much left now :-) (+1) – robjohn Apr 20 '13 at 9:40

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.

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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 $$0<a\frac{a_N!}{b}-\sum_{n=1}^{N}\frac{a_N!}{a_n!}\leq a_N!\sum_{k=1+a_N}^{\infty}\frac{1}{k!} <a_N!\frac{2}{(1+a_N)!}=\frac{2}{1+a_N}\leq 1.$$

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I don't understand why your generalization applies here, since $(n^2)! \neq n!^2$. – Martin Brandenburg Apr 22 '13 at 8:40
You are quite right. I will erase that comment. And this raises the question of irrationality of $\sum_n (n!)^{-a_n}$ :) – Captain Darling Apr 22 '13 at 11:48

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