Compact Proof that the constant $\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k.k!}$ is irrational

Question

In an answer some time ago I referenced a short proof that the constant $e$ is irrational by A. R. G. MacDivitt and Yukio Yanagisawa (The Mathematical Gazette , Volume 71 , Issue 457 , October 1987 , pp. 217)

What is the shortest proof that there is an irrational number?

I think the technique can be applied to prove that the constant $U_1=\sum_{k=1}^\infty \frac{1}{k.k!}$ is irrational.

Step 1: Suppose $U_1$ is rational; $U_1$ being defined by the infinite sum, $\sum_{k=1}^\infty \frac{1}{k.k!}$; that is:

$$\sum_{k=1}^\infty \frac{1}{k.k!}=\frac{m}{n}\tag{1},$$

where $m$ and $n$ are positive integers.

Step 2: Multiply both sides of (1) by $D(H_n).n!$, where $D(H_n)$ is the denominator of the Harmonic Number $n$ (simplified form for neatness, although this choice is not essential to the proof) thus

$$D(H_n).n!\left(1+\frac{1}{1!}+\frac{1}{2\times 2!}+\frac{1}{3\times 3!}+...+\frac{1}{n.n!} \right) \\+\frac{D(H_n)}{(n+1)^2}+\frac{D(H_n)}{(n+1)(n+2)^2}+\frac{D(H_n)}{(n+1)(n+2)(n+3)^2}+...=D(H_n)n!\,\frac{m}{n}\tag{2}$$

so that $S=\frac{D(H_n)}{(n+1)^2}+\frac{D(H_n)}{(n+1)(n+2)^2}+...$ is a positive integer, with $nS \ge \frac{D(H_n)}{(n+1)}$.

Step 3: Construct contradiction.

$nS\ge \frac{D(H_n)}{(n+1)}$ can be written $(n+1)S-\frac{D(H_n)}{(n+1)} \ge S$, implying that

$$\frac{D(H_n)}{(n+2)^2}+\frac{D(H_n)}{(n+2)(n+3)^2}+...\ge \frac{D(H_n)}{(n+1)^2}+\frac{D(H_n)}{(n+1)(n+2)^2}+...;$$

which after comparing both sides of the inequality term by term can immediately seen to be a contradiction. (Assuming I haven't made a mistake)

Assuming this is correct I am not sure if the same method can apply to the alternating sum $$U_2=\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k.k!}$$

As the sum $U_2$ is unconditionally convergent, adjacent positive and negative terms in the sum can be brought together giving positive terms only, thus

$$U_2=\sum_{k=1}^\infty \frac{(2k-1)^2+2k}{(2k-1)2k(2k)!}$$

but unfortunately this gives a numerator depending on $k$ and I think this stops the same proof from working.

For added background

$U_1$ and $U_2$ are related to the Euler-Mascheroni Constant, $\gamma$, thus

$$U_1=-\gamma+Ei(1)$$ $$U_2=\gamma-Ei(-1)$$

where $Ei(x)$ is exponential integral.

The Euler-Gompertz constant $\delta$ is $$\delta=-e\,Ei(-1)=eU_2-e\gamma$$

Additional Comment

It should also be noted that

$$U_1=\sum_{k=1}^\infty \frac{1}{(2k-1)(2k-1)!}+\sum_{k=1}^\infty \frac{1}{(2k)(2k)!}$$

and

$$U_2=\sum_{k=1}^\infty \frac{1}{(2k-1)(2k-1)!}-\sum_{k=1}^\infty \frac{1}{(2k)(2k)!}$$

Therefore $U_2$ the number we are trying to prove irrational is equal to

$$U_2=U_1 -2\sum_{k=1}^\infty \frac{1}{(2k)(2k)!}$$

which shows how infuriatingly elusive irrationality proofs can be.

Additional Comment 09/04/2024

One conjectured identity for $U_2$ involving only positive terms that I have found is

$$U_2=\sum_{k=1}^\infty \frac{1}{(2k)!}+\sum_{k=1}^\infty \frac{1}{\binom{2k-2}{k-1}\left(2\,k! (2k-1)\right)^2}$$

which is

$$U_2=\cosh(1)-1+\sum_{k=1}^\infty \frac{1}{\binom{2k-2}{k-1}\left(2\,k! (2k-1)\right)^2}$$

Mathematica verifies it, but I am not sure how to prove it though.

Answer 1

An important step in the above proof appears to be missing. Specifically, no justification is offered for the inequality $nS≥D(H_n)/(n+1)$. The corresponding step in the proof of MacDivitt and Yanagisawa (1987) -- i.e., $nS≥1$ -- follows immediately from the fact that $S$ must be a positive integer. However, the positive-integer status of $S$ is not explicitly used in the current proof. (Please correct me if I've overlooked something.)

Furthermore, direct numerical computation of $S$ for large values of $n$ (e.g., using WolframAlpha) suggests $nS>D(H_n)/(n+2)$, whereas $nS<D(H_n)/(n+1)$ follows from approximating $S$ by an infinite geometric series with common ratio $1/(n+1)$.

Answer 2

I believe the identity in the addendum of April 9, 2024 can be shown as follows: $$U_{2}=\sum_{k=1}^{\infty}\dfrac{1}{\left(2k-1\right)\left(2k-1\right)!}-\sum_{k=1}^{\infty}\dfrac{1}{\left(2k\right)\left(2k\right)!}$$ $$=\sum_{k=1}^{\infty}\dfrac{1}{\left(2k-1\right)^{2}\left(2k-2\right)!}-\sum_{k=1}^{\infty}\dfrac{1}{\left(2k\right)^{2}\left(2k-1\right)\left(2k-2\right)!}$$ $$=\sum_{k=1}^{\infty}\dfrac{\left(2k\right)^{2}-\left(2k-1\right)}{\left(2k-1\right)^{2}\left(2k\right)^{2}\left(2k-2\right)!}$$ $$=\sum_{k=1}^{\infty}\dfrac{\left(2k\right)^{2}-2k}{\left(2k-1\right)^{2}\left(2k\right)^{2}\left(2k-2\right)!}+\sum_{k=1}^{\infty}\dfrac{1}{\left(2k-1\right)^{2}\left(2k\right)^{2}\left(2k-2\right)!}$$ $$=\sum_{k=1}^{\infty}\dfrac{1}{\left(2k-1\right)\left(2k\right)\left(2k-2\right)!}+\sum_{k=1}^{\infty}\dfrac{\left[\left(k-1\right)!\right]^{2}}{\left(2k-1\right)^{2}\left(2k!\right)^{2}\left(2k-2\right)!}$$ $$=\sum_{k=1}^{\infty}\dfrac{1}{\left(2k\right)!}+\sum_{k=1}^{\infty}\dfrac{1}{\tbinom{2k-2}{k-1}\left(2k-1\right)^{2}\left(2k!\right)^{2}}.$$