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,

• Why would it be allowed to change the order of the limits? Commented Feb 20, 2023 at 20:53
• I don't think the series converge. If you have $\sum 1/n^s$ with some $s<1$, the sum will diverge so what allow us here to say that $\sum (-1)^{n-1}/n^s$ converges?
– tac
Commented Feb 21, 2023 at 0:20
• @tac The sum converges for $s>0$ ($\Re(s)>0$ if $s$ is a complex number).
– user
Commented Feb 28, 2023 at 8:39
• To keep things simple, we know that $$\zeta(0) = - \frac{1}{2}$$ And $$\eta(s) = (1 - 2^{1-s}) \, \zeta(s) \hspace{5mm} \cdots (1)$$ Now if I put 0 in (1) we get $$\eta(0) = \frac{1}{2}$$ Commented Feb 28, 2023 at 9:52
• I was not asking a question, I was showing to you why your calculation isn't a proof. Commented Mar 14, 2023 at 19:36

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 $$\lim_{s \to 0^+} Z(s)$$?

(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.

• Thank you David .. very helpful answer (+1). Commented Mar 1, 2023 at 19:06

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

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

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.

• My answer is right based on David's answer Commented Mar 1, 2023 at 23:45
• @AliShadhar From what I understand, the proof of $\eta(0)=\frac{1}{2}=\underset{s\to 0^+}{\text{lim}}\left(\underset{{N\to\infty}}{\text{lim}}\left(\sum\limits_{n=1}^N\frac{(-1)^{\,n-1}}{n^s}\right)\right)$ follows from the proof of $f(1)=\frac{1}{2}=\underset{x\to 1^-}{\text{lim}}\left(\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N (-1)^{\,n-1}\, x^n\right)\right)$ where the analytic continuation of $f(x)$ is $\frac{x}{x+1}$. Commented Mar 2, 2023 at 3:56
• @AliShadhar You used $g(1)=\frac{1}{2}=\underset{x\to 1^-}{\text{lim}}\left(\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N (-x)^{\,n-1}\right)\right)$ which has the analytic continuation $\frac{1}{x+1}$ rather than f(x). Commented Mar 2, 2023 at 3:56
• Thank you Steven Commented Mar 2, 2023 at 4:09
• @AliShadhar As I understand it, you used the analytic continuation of the Maclaurin series for $g(x)=\frac{1}{x+1}$ as $x\to 1^-$ to derive the analytic continuation of the Dirichlet series for $\eta(s)$ as $s\to 0^+$. I believe the techniques illustrated in my answer above to extend the convergence of the Dirichlet series for $\eta(s)$ from $\Re(s)>0$ to $s\in\mathbb{C}$ can also be used to extend the convergence of the Maclaurin series for $f(x)=\frac{x}{x+1}$ and $g(x)=\frac{1}{x+1}$ from $|x|<1$ to $\Re(x)>-1$. See my related question at math.stackexchange.com/q/4652009. Commented Mar 4, 2023 at 16:32

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

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