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Let's define the functions $$A_n(q)=\sum_{k=0}^n(-1)^k\cdot\frac{(1+q)q^k}{1+q^{2k+1}}\cdot\frac{2k+1}{n+k+1}\binom{2n}{n-k}.$$

I'm interested in the following:

QUESTION. Let $n\geq1$ be integers. Are these limiting values non-negative integers? $$\lim_{q\rightarrow1}\frac{A_n(q)}{(1-q)^{2n}}$$

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    $\begingroup$ The sequence $(a_n)$, where $a_n:=\lim_{q\to1}\frac{A_n(q)}{(1-q)^{2n}}$, appears to coincide with oeis.org/A000364 $\endgroup$
    – Iosif Pinelis
    Commented Feb 3, 2023 at 1:15

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If we expand around $q=1$ $$\frac{A_n(q)}{(1-q)^{2n}}=\sum_{k=0}^\infty a_k\,(1-q)^n$$
as @Iosif Pinelis already commented, the $a_0$ coreespond to sequence $A000364 $ in $OEIS$.

An interesting formulation given in the formula section is $$\color{blue}{a_0=(-1)^n \,2^{4 n+1}\, \left(\zeta \left(-2 n,\frac{1}{4}\right)-\zeta \left(-2 n,\frac{3}{4}\right)\right)}$$ and notice that $\color{blue}{a_1=n \,a_0}$

I have not been able to identify the sequence for $a_2$. We can notice that they are all negative which then give a simple upper bound.

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