This question is about the relationship between the known globally convergent formula illustrated in formula (1) below and the conjectured formula illustrated in formula (2) below which I originally defined in formula (11) of my related question.
$$\quad \eta(s)=\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{n=0}^K\frac{1}{2^{n+1}}\sum\limits_{k=0}^n\frac{(-1)^k}{(k+1)^s}\binom{n}{k}\right)\tag{1}$$
$$\quad \eta(s)=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^{K+1}}\sum\limits_{n=0}^K\frac{(-1)^n}{(n+1)^s}\sum\limits_{k=0}^{K-n}\binom{K+1}{K-n-k}\right)\tag{2}$$
I verified formulas (1) and (2) above evaluate exactly the same for $K=0$ and the first $100$ positive integer values of $K$, so this seems to suggest a relationship between the two formulas above.
Question (1): Can the following relationship be proven true for all integer values of $K\ge 0$?
$$\quad\frac{1}{2^{K+1}}\sum\limits_{n=0}^K\frac{(-1)^n}{(n+1)^s}\sum\limits_{k=0}^{K-n}\binom{K+1}{K-n-k}=\sum\limits_{n=0}^K\frac{1}{2^{n+1}}\sum\limits_{k=0}^n\frac{(-1)^k}{(k+1)^s}\binom{n}{k}\tag{3}$$
I've noticed formulas (1) and (2) for $\eta(s)$ above seem to evaluate exactly correct when s is a non-positive integer and |s|≤K.
Question (2): Does a proof of the global convergence of formulas (1) and (2) above imply anything about the exact convergence of these formulas when $s$ is a non-positive integer and $|s|\le K$?
Question (3): Does a proof the exact convergence of formulas (1) and (2) above when $s$ is a non-positive integer and $|s|\le K$ imply anything about the global convergence of these formulas?
June 22, 2023 Update:
Note formulas (1) and (2) above for the Dirichlet eta function $\eta(s)$ can also be evaluated as defined in formulas (4) and (5) below which I believe are exactly equivalent for all integer values of $K$ where $_2F_1(a,b;c;z)$ is a hypergeometric function and $P_n^{(\alpha,\beta)}(x)$ is the Jacobi Polynomial.
$$\eta(s)=\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^K \frac{1}{2^n} \sum\limits_{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{k^s}\right),\quad s\in\mathbb{C}\tag{4}$$
$$\eta(s)=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^K} \sum\limits_{n=1}^K \frac{(-1)^{n-1}}{n^s} \sum\limits_{k=0}^{K-n} \binom{K}{K-n-k}\right)$$ $$=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^K} \sum\limits_{n=1}^K \frac{(-1)^{n-1}}{n^s}\, \binom{K}{K-n} \, _2F_1(1,n-K;n+1;-1)\right)$$ $$=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^K} \sum\limits_{n=1}^K \frac{(-1)^{n-1}}{n^s}\, P_{K-n}^{(n,-K)}(3)\right),\quad s\in\mathbb{C}\tag{5}$$
This answer to one of my related questions on MathOverflow proves the equivalence of formulas (6) and (7) below for $\frac{x}{x+1}$ which are derived from formulas (4) and (5) above for $\eta(s)$ by the mappings $\frac{1}{k^s}\to x^k$ and $\frac{1}{n^s}\to x^n$.
$$\frac{x}{x+1}=\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^K \frac{1}{2^n} \sum\limits_{k=1}^n (-1)^{k-1} \binom{n-1}{k-1}\, x^k\right),\quad|1-x|<2\tag{6}$$
$$\frac{x}{x+1}=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^K} \sum\limits_{n=1}^K (-1)^{n-1}\, x^n \sum\limits_{k=0}^{K-n} \binom{K}{K-n-k}\right)$$ $$=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^K} \sum\limits_{n=1}^K (-1)^{n-1} \binom{K}{K-n} \, _2F_1(1,n-K;n+1;-1)\, x^n\right)$$ $$=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^K}\sum\limits_{n=1}^K (-1)^{n-1}\, P_{K-n}^{(n,-K)}(3)\, x^n\right),\quad|1-x|<2\tag{7}$$
I believe formulas (6) and (7) above are exactly equivalent to formula (8) below.
$$\frac{x}{x+1}=\underset{K\to\infty}{\text{lim}}\left(\frac{x-2^{-K}\, x\, (1-x)^K}{x+1}\right),\quad|1-x|<2\tag{8}$$
The referenced proof of formula (7) above involves a step where the order of summation is reversed, so perhaps an analogous approach can be used to prove the equivalence of formulas (4) and (5) above which are equivalent to formulas (1) and (2) above.