The question here is motivated by this Math StackExchange question and this Math Overflow question which indicate the evaluation of the Dirchleta eta function
$$\eta(s)=\underset{K\to\infty}{\text{lim}}\left(\sum_{n=1}^K \frac{(-1)^{\,n-1}}{n^s}\right),\quad\Re(s)>0\tag{1}$$
as $s\to 0^+$ is related to the evaluation of Maclaurin series such as
$$\frac{x}{x+1}=\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^K (-1)^{\,n-1}\, x^n\right),\quad |x|<1\tag{2}$$
as $x\to 1^-$, and therefore the analytic continuation of formula (1) for $\eta(s)$ above at $s=0$ must be equal to $\frac{x}{x+1}$ evaluated at $x=1$ (i.e. $\eta(0)=\frac{1}{2}$).
Now consider the following two globally convergent formulas for the Dirichlet eta function $\eta(s)$ 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.
$$\eta(s)=\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^K \frac{1}{2^n} \sum\limits_{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{k^s}\right),\quad s\in\mathbb{C}\tag{3}$$
$$\eta(s)=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^K} \sum\limits_{n=1}^K \frac{(-1)^{n-1}}{n^s} \sum\limits_{k=0}^{K-n} \binom{K}{K-n-k}\right)$$ $$=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^K} \sum\limits_{n=1}^K \frac{(-1)^{n-1}}{n^s}\, \binom{K}{K-n} \, _2F_1(1,n-K;n+1;-1)\right)$$ $$=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^K} \sum\limits_{n=1}^K \frac{(-1)^{n-1}}{n^s}\, P_{K-n}^{(n,-K)}(3)\right),\quad s\in\mathbb{C}\tag{4}$$
The conjectured formulas (5) and (6) below for $\frac{x}{x+1}$ are derived from formulas (3) and (4) above for $\eta(s)$ by the mappings $\frac{1}{k^s}\to x^k$ and $\frac{1}{n^s}\to x^n$.
$$\frac{x}{x+1}=\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^K \frac{1}{2^n} \sum\limits_{k=1}^n (-1)^{k-1} \binom{n-1}{k-1}\, x^k\right),\quad\Re(x)>-1\tag{5}$$
$$\frac{x}{x+1}=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^K} \sum\limits_{n=1}^K (-1)^{n-1}\, x^n \sum\limits_{k=0}^{K-n} \binom{K}{K-n-k}\right)$$ $$=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^K} \sum\limits_{n=1}^K (-1)^{n-1} \binom{K}{K-n} \, _2F_1(1,n-K;n+1;-1)\, x^n\right)$$ $$=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^K}\sum\limits_{n=1}^K (-1)^{n-1}\, P_{K-n}^{(n,-K)}(3)\, x^n\right),\quad\Re(x)>-1\tag{6}$$
My initial intuition was that formulas (5) and (6) above extend the range of convergence from $|x|<1$ related to formula (2) above to $\Re(x)>-1$, but the diverging oscillation in the right-half plane illustrated in Figure (1) following my question below made be begin to seriously question my initial intuition. I began to wonder if formulas (5) and (6) perhaps only converge for $|x|<r$ for some $r>1$ with the additional constraint $\Re(x)>1$, or perhaps only converge similar to
$$\frac{x}{x+1}=\frac{1}{2} \underset{K\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^K (-1)^{n-1}\, x^n+\sum\limits_{n=1}^{K+1} (-1)^{n-1} x^n\right)$$ $$=\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^K (-1)^{n-1}\, x^n+\frac{1}{2}\, (-1)^K\, x^{K+1}\right)\tag{7}$$
which I believe converges for $|x|\le 1\land x\ne -1$.
But then I remembered that formulas (3) and (4) for $\eta(s)$ above exhibit a similar diverging oscillation when evaluated in the left-half plane which is illustrated in Figure (2) following my question below, so perhaps the diverging oscillations in the evaluations of formulas (5) and (6) above for $\frac{x}{x+1}$ in the right-half plane illustrated in Figure (1) below are related to the precision of the evaluations.
Question: Is it true that formulas (5) and (6) for $\frac{x}{x+1}$ above converge for $\Re(x)>-1$? If not, what is the convergence of formulas (5) and (6) above?
Figure (1) below illustrates formula (6) above evaluated at $K=100$ in orange and $K=200$ in green overlaid on the blue reference function $\frac{x}{x+1}$. Note the diverging oscillation in the evaluation of formula (6) seems to occur earlier (closer to the origin) as the evaluation limit $K$ increases.
Figure (1): Illustration of formula (6) (orange and green) overlaid on the blue reference function $\frac{x}{x+1}$
Figure (2) below illustrates formula (4) above evaluated at $K=75$ in orange and $K=150$ in green overlaid on the blue reference function $\eta(s)$. Note the diverging oscillation in the evaluation of formula (4) seems to occur earlier (closer to the origin) as the evaluation limit $K$ increases.
Figure (2): Illustration of formula (4) (orange and green) overlaid on the blue reference function $\eta(s)$
Now consider the series
$$\frac{1}{x+1}=\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^K \frac{(-1)^{\,n-1}}{x^n}\right),\quad |x|>1.\tag{8}$$
The conjectured convergence $\Re(x)>-1$ of formulas (5) and (6) for $\frac{x}{x+1}$ above implies formulas (9) and (10) below extend the convergence from $|x|>1$ related to formula (8) for $\frac{1}{x+1}$ above to $\Re(x)<-1\lor\Re(x)>0$.
$$\frac{1}{x+1}=\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^K \frac{1}{2^n} \sum\limits_{k=1}^n (-1)^{k-1} \binom{n-1}{k-1}\, \frac{1}{x^k}\right),\quad\Re(x)<-1\lor\Re(x)>0\tag{9}$$
$$\frac{1}{x+1}=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^K} \sum\limits_{n=1}^K (-1)^{n-1}\, \frac{1}{x^n} \sum\limits_{k=0}^{K-n} \binom{K}{K-n-k}\right)$$ $$=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^K} \sum\limits_{n=1}^K (-1)^{n-1} \binom{K}{K-n} \, _2F_1(1,n-K;n+1;-1)\, \frac{1}{x^n}\right)$$ $$=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^K}\sum\limits_{n=1}^K (-1)^{n-1}\, P_{K-n}^{(n,-K)}(3)\, \frac{1}{x^n}\right),\quad\Re(x)<-1\lor\Re(x)>0\tag{10}$$
Figure (3) below illustrates formulas (8) and (10) above in orange and green overlaid on the blue reference function $\frac{1}{x+1}$ where formulas (8) and (10) are both evaluated at $K=100$. Note the evaluation of formula (10) in Figure (3) below illustrates a diverging oscillation similar to the diverging oscillations seen in Figures (1) and (2) above which is perhaps related to the precision of the evaluation.
Figure (3): Illustration of formulas (8) and (10) (orange and green) overlaid on the blue reference function $\frac{1}{x+1}$