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

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.

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 $$x\to 1^-$$.

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.

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

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

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 $$\frac{1}{n^s}\to x^n$$.

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

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

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?

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 $$|1-x|<2$$.
Details on \eqref{6}: \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}
• With respect to (6), I believe you're saying $$\sum\limits_{k=0}^{K-n} \binom{K}{K-n-k}=\sum\limits_{j=0}^{K-n} \binom{K}{j}$$ but I don't understand how you interchange the order of summation in $$\frac{1}{2^K} \sum\limits_{n=1}^K (-1)^{n-1} x^n \sum\limits_{j=0}^{K-n} \binom{K}{j}$$ since the inner sum is still a function of $n$. May 23, 2023 at 17:04
• After seeing your answer, I was hoping it would initially lead to a proof of the equivalence of formulas $\eqref{5}$ and $\eqref{6}$ and also perhaps ultimately lead to a proof of the equivalence of formulas $\eqref{3}$ and $\eqref{4}$. With respect to the latter, please see my related Math StackExchange and Math Overflow questions. May 23, 2023 at 17:56