Is my proof of $\eta(0)=\frac{1}{2}$ correct?

Question

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,

Answer 1

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.

Answer 2

You need to perform an analytic extension of the Dirichlet eta function

$$\eta(s)=\lim_{N\to\infty}\left(\sum_{n=1}^N\frac{(-1)^{\,n-1}}{n^s}\right),\quad\Re(s)>0.\tag{1}$$

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One way to do this is

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

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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 $N\to\infty$.

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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.

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

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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.

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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.

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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.

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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).

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

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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.

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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 $n$.

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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.

Answer 3

For $|x|<1$ no problem to say that $$(1+x)\sum_{n\ge 1} (-x)^{n-1} n^{-s} =1 +\sum_{n\ge 2} (-x)^{n-1} (n^{-s}-(n-1)^{-s})$$ $$(1+x)^2\sum_{n\ge 1} (-x)^{n-1} n^{-s} =1+x-2^{-s}x+\sum_{n\ge 3} (-x)^{n-1} (n^{-s}-2(n-1)^{-s}+(n-2)^{-s})$$

The point is that the latter series converges absolutely and uniformly for $s\in [0,1], x\in [0, 1]$. So we can change the order of limits and obtain that

$$4\eta(0)=\lim_{s\to 0^+}\lim_{x\to 1^-} (1+x)^2\sum_{n\ge 1} (-x)^{n-1} n^{-s}= \lim_{x\to 1^-}\lim_{s\to 0^+} (1+x)^2\sum_{n\ge 1} (-x)^{n-1} n^{-s}$$ $$= \lim_{x\to 1^-} (1+x)^2\sum_{n\ge 1} (-x)^{n-1}=\lim_{x\to 1^-}\frac{(1+x)^2}{1+x} = 2$$