Is this a valid method of extending convergence of the Maclaurin series for $\frac{x}{x+1}$ from $|x|<1$ to $\Re(x)>-1$?

Question

I originally asked this question on Math StackExchange a few months ago and no answers or even comments have yet been posted, so I'm asking this question again here on Math OverFlow.

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This Math StackExchange question and this Math Overflow question indicate the evaluation of the Dirchleta eta function

$$\eta(s)=\underset{K\to\infty}{\text{lim}}\left(\sum_{n=1}^K \frac{(-1)^{\,n-1}}{n^s}\right),\quad\Re(s)>0\label{1}\tag{1}$$

as $s\to 0^+$ is related to the evaluation of Maclaurin series such as

$$\frac{x}{x+1}=\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^K (-1)^{\,n-1}\, x^n\right),\quad |x|<1\label{2}\tag{2}$$

as $x\to 1^-$.

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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.

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$$\eta(s)=\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^K \frac{1}{2^n} \sum\limits_{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{k^s}\right),\quad s\in\mathbb{C}\label{3}\tag{3}$$

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$$\eta(s)=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^K} \sum\limits_{n=1}^K \frac{(-1)^{n-1}}{n^s} \sum\limits_{k=0}^{K-n} \binom{K}{K-n-k}\right)$$ $$=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^K} \sum\limits_{n=1}^K \frac{(-1)^{n-1}}{n^s}\, \binom{K}{K-n} \, _2F_1(1,n-K;n+1;-1)\right)$$ $$=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^K} \sum\limits_{n=1}^K \frac{(-1)^{n-1}}{n^s}\, P_{K-n}^{(n,-K)}(3)\right),\quad s\in\mathbb{C}\label{4}\tag{4}$$

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The conjectured formulas \eqref{5} and \eqref{6} below for $\frac{x}{x+1}$ are derived from formulas \eqref{3} and \eqref{4} above for $\eta(s)$ by the mappings $\frac{1}{k^s}\to x^k$ and $\frac{1}{n^s}\to x^n$.

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$$\frac{x}{x+1}=\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^K \frac{1}{2^n} \sum\limits_{k=1}^n (-1)^{k-1} \binom{n-1}{k-1}\, x^k\right),\quad\Re(x)>-1\label{5}\tag{5}$$

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$$\frac{x}{x+1}=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^K} \sum\limits_{n=1}^K (-1)^{n-1}\, x^n \sum\limits_{k=0}^{K-n} \binom{K}{K-n-k}\right)$$ $$=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^K} \sum\limits_{n=1}^K (-1)^{n-1} \binom{K}{K-n} \, _2F_1(1,n-K;n+1;-1)\, x^n\right)$$ $$=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^K}\sum\limits_{n=1}^K (-1)^{n-1}\, P_{K-n}^{(n,-K)}(3)\, x^n\right),\quad\Re(x)>-1\label{6}\tag{6}$$

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Question: Is it true that formulas \eqref{5} and \eqref{6} for $\frac{x}{x+1}$ above converge for $\Re(x)>-1$? If not, what is the convergence of formulas \eqref{5} and \eqref{6} above?

Answer 1

In \eqref{5}, the inner sum is obviously $x(1-x)^{n-1}$. So, the limit in \eqref{5} (equal $\dfrac x{x+1}$ indeed) exists if and only if $|1-x|<2$.

In \eqref{6}, using the substitution $k=K-n-j$ in the inner sum and then interchanging the order of summation, we see that the iterated sum under the limit sign is $\Big(1-\Big(\dfrac{1-x}2\Big)^K\Big)\dfrac x{x+1}$. So, the limit in \eqref{6} (equal $\dfrac x{x+1}$ indeed) exists if and only if $|1-x|<2$.

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Details on \eqref{6}: \begin{equation} \begin{aligned} &\frac1{2^K} \sum_{n=1}^K (-1)^{n-1}\, x^n \sum_{k=0}^{K-n} \binom K{K-n-k} \\ &=\frac1{2^K} \sum_{n=1}^K (-1)^{n-1}\, x^n \sum_{j=0}^{K-n} \binom Kj \\ &=\frac1{2^K} \sum_{j=0}^{K-1} \binom Kj \sum_{n=1}^{K-j} (-1)^{n-1}\, x^n \\ &=\frac1{2^K} \sum_{j=0}^{K-1} \binom Kj \frac x{1+x}\,(1-(-x)^{K-j}) \\ &=\frac1{2^K}\,\frac x{1+x}\, \sum_{j=0}^{K-1} \binom Kj (1-(-x)^{K-j}) \\ &=\frac1{2^K}\,\frac x{1+x}\, \sum_{j=0}^K \binom Kj (1-(-x)^{K-j}) \\ &=\frac1{2^K}\,\frac x{1+x}\, (2^K-(1-x)^K) \\ &=\frac x{1+x}\,\Big(1-\Big(\dfrac{1-x}2\Big)^K\Big). \end{aligned} \end{equation}