I originally asked this question on [Math StackExchange](https://math.stackexchange.com/q/4652009) 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](https://math.stackexchange.com/q/4643164) and [this Math Overflow question](https://mathoverflow.net/q/19410) 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](https://en.wikipedia.org/wiki/Hypergeometric_function) and $P\_n^{(\alpha,\beta)}(x)$ is the [Jacobi Polynomial](https://mathworld.wolfram.com/JacobiPolynomial.html).




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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?
> 
> 
> 



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


