Why is $\sum_{n\ge1}\frac{\text B_\frac n2(-n,n)}{n^a b^n }=-\sum_{n\ge1}\frac{\left(\frac 2n-1\right)^n}{n^{a+1}b^n}$ close to (reciprocal) integers?

Question

Here is a possible closed form of a sum with tetration in it made by equating coefficients of the Incomplete Beta function. This question is inspired by:

Closed form of $$\sum\limits_{n=1}^\infty \frac{γ(n+a,bn)n^k}{(n+c)!}$$ with the Lower Gamma function?

Where we actually found many closed forms. Here is another simple one where there are 2 variations with and without an $n!$ in the denominator.

$$\text B_z(a,b)=\int_0^z t^{a-1}(1-t)^{b-1}dt\implies \sum_{n\ge1}\text B_\frac n2(-n,n)=-\sum_{n\ge1}\frac{\left(\frac 2n-1\right)^n}n=-0.995447…, \sum_{n\ge1} \frac{\text B_\frac n2(-n,n)}{nn!}=-\sum_{n\ge1}\frac{\left(\frac2n-1\right)^n}{n^3n!^2}=-0.9999635… $$

The reason for this approximation is probably due to the numerator of the sum being much smaller than the denominator, but this is not the case with $$\sum_{n\ge1}\text B_\frac n2(-n,n)=-\sum_{n\ge1}\frac{\left(\frac 2n-1\right)^n}n=-0.995447… $$

which oscillated between values below and above $-1$ with the partial sums and converges slower than $\sum\limits_{n\ge1}\frac 1{n^2}$.

Another example which has an imaginary part greater than $1$:

$$\sum_{n\ge1} \text B_\frac n2 (-n,n)i^n=-\sum_{n\ge 1} \frac{(-1)^n (2-n)^n}{n^{n+1}}=-0.0078824-1.004251…i$$

which converges pretty slowly with this summand graph:

Of course

$$\lim_{a\to\infty}\sum_{n=1}^\infty \frac{\text B_\frac n2(-n,n)}{n^a}=-1$$

The question seems simple enough. You may also notice that for a mix of $a,b$:

$$\sum_{n\ge1}\frac{\text B_\frac n2(-n,n)n^{1+2i}}{2^n}=-\sum_{n\ge1}\left(\frac1 n-\frac12\right)^n n^{2i}=-0.50064…+0.00244…i≈-\frac12$$

Or

$$\sum_{n\ge1}\frac{3^n \text B_\frac n2(-n,n)}{n!}=\sum_{n\ge1} \frac{\left(\frac 6n-3\right)^n}{nn!}=-2.97616…≈-3$$

The question is why sums in the form of:

$$\sum_{n\ge1}\frac{\text B_\frac n2(-n,n)}{n^a b^n n!^{\{0\text{ or }1\}}}=-\sum_{n\ge1}\frac{\left(\frac 2n-1\right)^n}{n^{a+1}b^n n!^{\{0\text{ or }1\}}} $$ are very close to integers or reciprocal integers like in the referenced question. The $\{0\text{ or }1\}$ just means $0$ or $1$ as two different cases with and without the factorial in the denominator.

General pattern:

$$\sum_{n\ge1}\frac{\text B_\frac n2(-n,n)}{b^n n!^{\{0\text{ or }1\}}}≈-\frac 1b$$

$$\sum_{n\ge1}\frac{\text B_\frac n2(-n,n)}{n^a n!^{\{0\text{ or }1\}}} ≈-1$$

Please correct me and give me feedback!