Euler-Mascheroni Constant

Can I do something like this:

Define $x=\lim_{n\rightarrow \infty} b (\gamma-\sum_{1\le k \le b}\frac{1}{k}+\ln b) b! n! (n-1)^{\frac{n(n+1)}{2}}$, assuming $n\in Z$

If $\gamma\in Q$ then $\exists a,b \in Z$ s.t. $\gamma=\frac{a}{b}$, $b\neq \pm 1$

Then

$x=a(b!n!(n-1)^{\frac{n(n+1)}{2}})-b(\sum_{1\le k \le b}\frac{1}{k}-\ln n )(b! n! (n-1)^{\frac{n(n+1)}{2}})$, which, after some evaluation, is an integer.

Consider the definition of $\gamma$

$x=b(\sum_{b+1\le k\le n}\frac{1}{k})(b!n!(n-1)^{\frac{n(n+1)}{2}})=b\frac{(huge\quad nominator)}{n!}(b!)^2 n! (n-1)^{\frac{n(n+1)}{2}}$ and it turns out to be an integer as well.

Can I conclude that the constant in rational?

Note: Typos may exist.

That didn't make any sense whatsoever.

Define $x=\lim_{n\rightarrow \infty} b (\gamma-\sum_{1\le k \le b}\frac{1}{k}+\ln b)\alpha$, assuming $n\in Z$ and $\alpha=b! n! (n-1)^{\frac{n(n+1)}{2}}$

You do realise that you have $\displaystyle n! (n-1)^{\frac{n(n+1)}{2}}$ as $n \to \infty$?

Intriguing, though we know what you meant.

I am too lazy to introduce a new polynomial or something like that.

You missed the point. The word is "numerator", not "nominator".

Tends to infinity, but we can't just say it is infinity, because it is never infinity.

What I am saying is that your limit does not make sense. That way, I can prove pretty much everything to be a rational number.

You mean it doesn't converge?

I took the limit because of the definition of the constant.

He means that your $x\to \infty$ so you cannot conclude that $x\in\mathbb{Z},$ because it isn't once we introduce the limit (which is required given how you are using that definition of $\gamma$

But at first I have already assumed $n\in Z$ as the first case to be considered. I am kind of tired at the moment to do the remaining maths.

As long as $n\in\mathbb{Z}$ you are fine. But the definition of $\gamma$ isn't for some fixed $n$, so if you want to exploit it in any viable way you must let $n\to\infty$ in which case $x$ is no longer an integer.

How about of $n\notin Z$, then $x\notin Z$ (Just a claim), can it do the same trick?

No, because then your definition of $\gamma$ would make absolutely no sense

Out of interest, was this attempt inspired by a famous proof of the irrationality of e?

But isn't the summation in $\gamma$ forces $n$ to be natural number?

It is an infinite series over $\mathbb{N}$. I was referring to the sum not making sense if n were not an integer.