Is my proof of $\eta(0)=\frac{1}{2}$ correct? Is my following proof of $\eta(0)=\frac{1}{2}$ correct?
$$\eta(0)=\lim_{s\to 0^{+}}\eta(s)=\lim_{s\to 0^{+}}\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}$$
$$=\lim_{s\to 0^{+}}\left(\lim_{x\to 1^{-}}\sum_{n=1}^\infty \frac{(-x)^{n-1}}{n^s}\right)$$
$$\{\text{change the order of the limits}\}$$
$$=\lim_{x\to 1^{-}}\left(\lim_{s\to 0^{+}}\sum_{n=1}^\infty \frac{(-x)^{n-1}}{n^s}\right)$$
$$=\lim_{x\to 1^{-}}\left(\sum_{n=1}^\infty (-x)^{n-1}\right)$$
$$=\lim_{x\to 1^{-}}\left(\frac{1}{1+x}\right)$$
$$=\frac{1}{2}$$
I let $s\to 0^{+}$ since $\displaystyle\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}$ is defined for $\Re(s)>0$ and I let $x\to 1^{-}$ since $\displaystyle\sum_{n=1}^\infty x^n$ is defined for $|x|<1$.
Thank you,
 A: You need to perform an analytic extension of the Dirichlet eta function
$$\eta(s)=\lim_{N\to\infty}\left(\sum_{n=1}^N\frac{(-1)^{\,n-1}}{n^s}\right),\quad\Re(s)>0.\tag{1}$$

One way to do this is
$$\eta(s)=\lim_{N\to\infty}\left(\frac{1}{2}\left(\sum_{n=1}^N\frac{(-1)^{\,n-1}}{n^s}+\sum_{n=1}^{N+1}\frac{(-1)^{\,n-1}}{n^s}\right)\right)$$ $$=\lim_{N\to\infty}\left(\sum_{n=1}^N\frac{(-1)^{\,n-1}}{n^s}+\frac{(-1)^{N}}{2\, (N+1)^s}\right),\quad\Re(s)>-1.\tag{2}$$

Formula (2) for $\eta(s)$ above converges to $\frac{1}{2}$ at $s=0$ for any integer $N\ge 0$, but only more generally converges for $Re(s)>-1$ as $N\to\infty$.

Figure (1) below illustrates formula (2) for $\eta(s)$ above overlaid on the blue reference function $\eta(s)$ where formula (2) is evaluated at $N=100$ (orange) and $N=100000$ (green). Note formula (2) converges closer to $s=-1$ as the evaluation limit increases.


Figure (1): Illustration of formula (2) for $\eta(s)$ evaluated at $N=100$ (orange) and $N=100000$ (green) overlaid on the blue reference function.

Figure (2) and (3) below illustrate the real and imaginary parts of formula (2) for $\eta(s)$ above in orange overlaid on the blue reference function $\eta(s)$ where both are evaluated at $s=-\frac{1}{2}+i\, t$ and where formula (2) is evaluated at $N=1000$. Note formula (2) evaluates so accurately in Figures (2) and (3) below that it essentially hides the underlying blue reference function.


Figure (2): Illustration of formula (2) for $\Re\left(\eta\left(-\frac{1}{2}+i\, t\right)\right)$ in orange overlaid on the blue reference function.


Figure (3): Illustration of formula (2) for $\Im\left(\eta\left(-\frac{1}{2}+i\, t\right)\right)$ in orange overlaid on the blue reference function.

I originally investigated iterative application of this technique of analytic extension of the Dirichlet eta function $\eta(s)$ in this question which eventually lead to derivation of the conjectured globally convergent formula
$$\quad \eta(s)=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^{K+1}}\sum\limits_{n=0}^K\frac{(-1)^n}{(n+1)^s}\sum\limits_{k=0}^{K-n}\binom{K+1}{K-n-k}\right),\quad s\in\mathbb{C}\tag{3}$$
which I posted in formula (11) of this question and which I believe is exactly equivalent  for all $s\in\mathbb{C}$ and all $K\in\mathbb{N}$ to the known globally convergent formula
$$\quad\eta(s)=\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{n=0}^K\frac{1}{2^{n+1}}\sum\limits_{k=0}^n\frac{(-1)^k\,\binom{n}{k}}{(k+1)^s}\right),\quad s\in\mathbb{C}\tag{4}$$
(see Dirichlet eta function - Numerical algorithms).

I've believe formulas (3) and (4) for $\eta(s)$ above evaluate exactly correct when $s$ is a non-positive integer and $|s|\le K$.

I asked about a formal proof of the equivalence of formulas (3) and (4) above in this question, but no one has yet posted an answer.

Note formula (3) above is more efficient than formula (4) above as formula (3) moves the exponentiation operation from the inner sum over $k$ to the outer sum over $n$.

Mathematica likes to simplify formula (3) above as
$$\eta(s)=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^{K+1}} \sum\limits_{n=0}^K \frac{(-1)^n}{(n+1)^s} \binom{K+1}{K-n} \, _2F_1(1,n-K;n+2;-1)\right),\quad s\in\mathbb{C}\tag{5}$$
where $_2F_1(a,b;c;z)$ is a hypergeometric function, but user agno pointed out in a comment on my related Math Overflow question that formula (3) above can also be simplified as
$$\eta(s)=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^{K+1}} \sum\limits_{n=0}^K \frac{(-1)^n}{(n+1)^s}\, P_{K-n}^{(n+1,-K-1)}(3)\right),\quad s\in\mathbb{C}\tag{6}$$
where $P_n^{(\alpha,\beta)}(x)$ is the Jacobi Polynomial.
A: This approach does point to a valid result, which we discussed over at Mathoverflow: Let $a_n$ be a sequence of complex numbers such that
$$F(x):= \sum a_n x^n \ \text{and}\ Z(s) = \sum a_n n^{-s}$$
converge for $0 \leq x < 1$ and $s>0$. What can we say about the limits $\lim_{x \to 1^{-}} F(x)$ and $\lim_{s \to 0^+} Z(s)$?
The answers are
(1) It is possible that $\lim_{s \to 0^+} Z(s)$ exists but $\lim_{x \to 1^{-}} F(x)$ doesn't. An example is $a_n = n^{-1} e^{i c\log n}$ for nonzero real $c$. See Gerald Edgar's answer.
(2) If $\lim_{x \to 1^{-}} F(x)$ exists, then so does $\lim_{s \to 0^+} Z(s)$, and they are equal. See Brad Rogers' answer.
The second point is the one that justifies your computation. As you can see, the proof is highly nontrivial.
