I have here the known series of natural logarthem.
$$\ln \left( \frac {1}{1-x} \right)=x+\frac {x^2}{2}+\frac {x^3}{3}+\frac {x^4}{4}+\frac {x^5}{5}+\cdots \tag 1$$
and
$$\ln(1+x)=x-\frac {x^2}{2}+\frac {x^3}{3}-\frac {x^4}{4}+\frac {x^5}{5}-\cdots \tag 2$$
Subtract $(2)$ from $(1)$
$$\ln\left(\frac {1}{1-x}\right)-\ln(1+x)=x^2 +\frac {x^4}{2}+\frac {x^6}{3}+\frac {x^8}{4}+\cdots \tag 3$$
Now if we let $x \to 1$
$$\lim_{x \to 1}\left[\ln\left(\frac 1{1-x}\right)-(x^2 +\frac {x^4}{2}+\frac {x^6}{3}+\frac {x^8}{4}+\cdots)\right]=\ln (2) \tag 4$$
So what is the difference between this and Euler–Mascheroni constant limit?
$$\gamma=\lim_{n \to\infty} \left(-\ln (n)+\sum_{k=1}^{n} \frac 1 k \right) \tag 5$$
I am sure that $\gamma \neq -\ln(2)$
There must be a problem that I can not recognize.
The problem is that you have four different quantities all diverging as $n\to\infty$ and $x\to1^-$. Whether the difference between any two of these diverges or converges to any particular value depends on exactly how each one diverges.
You can make $\ln \left( \frac {1}{1-x} \right)$ and $ln(n)$ diverge in exactly the same way by setting them equal, that is, by setting $x = 1 - \frac1n$. Now compare $\sum_{k=1}^{n} \frac 1 k$ with $x^2 +\frac {x^4}{2}+\frac {x^6}{3}+\frac {x^8}{4}+\cdots$ for various increasing values of $n$. You may notice that when $n = 1$, $\sum_{k=1}^{n} \frac 1 k - \left(x^2 +\frac {x^4}{2}+\frac {x^6}{3}+\frac {x^8}{4}+\cdots\right) = 1$, and that for larger values of $n$ this difference does not approach zero but rather continues to grow.
This "excess" value of $\sum_{k=1}^{n} \frac 1 k$ explains why $\gamma > -\ln(2)$.