# Question on conjectured formulas for $B_n$, $\eta(s)$, and $\zeta(s)$

The $$B(s)$$ function defined below is the extension of the Bernoulli number function $$B_n$$ in the spirit of Analytic Continuation of Bernoulli Numbers ... except the definition here follows the convention $$B_1=-1/2$$ instead of $$B_1=1/2$$ which is advocated by the referenced paper. Note $$B(n)=B_n$$ when n is a positive integer.

(1) $$\quad B(s)=(-1)^{s-1}\,s\,\zeta(1-s)$$

The following two figures illustrate the real and imaginary parts of the $$B(s)$$ function defined in formula (1) above. The red discrete portion of Figure (1) illustrates the evaluation of the real part of $$B(s)$$ defined in formula (1) where $$s=n$$ and n is a positive integer, and the red discrete portion of Figure (2) illustrates the evaluation of the imaginary part of $$B(s)$$ defined in formula (1) at integer values of $$s$$. Note the imaginary part of $$B(s)$$ always seem to evaluate to zero at negative as well as positive integers. Also note $$B(0)$$ as defined in formula (1) above is indeterminate but seems to converge to $$B_0=1$$ in a limit sense.

Figure (1): Illustration of real part of $$B(s)$$

Figure (2): Illustration of imaginary part of $$B(s)$$

The following conjectured formulas are refinements of formulas defined in my three previous questions on the Dirichlet eta function, Riemann zeta function, and Bernoulli number function.

(2) $$\quad \eta(s)=\underset{m\to\infty}{\text{lim}}\left(\frac{1}{2^m}\sum\limits_{n=1}^m\frac{(-1)^{n-1}}{n^s}\sum\limits_{i=0}^{m-n}\binom{m}{m-n-i}\right)$$

(3) $$\quad \zeta(s)=\underset{m\to\infty}{\text{lim}}\left(\frac{1}{2^m\left(1-2^{1-s}\right)}\sum\limits_{n=1}^m\frac{(-1)^{n-1}}{n^s}\sum\limits_{i=0}^{m-n}\binom{m}{m-n-i}\right)\,,\quad s\ne 1+\frac{2\,π\,i\,k}{\log(2)}$$

(4) $$\quad B(s)=\underset{m\to\infty}{\text{lim}}\left(\frac{(-1)^{s-1}\,s}{2^m\,\left(1-2^s\right)}\sum\limits_{n=1}^m (-1)^{n-1} n^{s-1}\sum\limits_{i=0}^{m-n}\binom{m}{m-n-i}\right)\,,\quad s\ne \frac{2\,π\,i\,k}{\log(2)}$$

Formulas (2), (3), and (4) above are related and a proof of any one of these would imply the correctness of all three formulas. Formulas (2), (3), and (4) above for $$\eta(s)$$, $$\zeta(s)$$ and $$B(s)$$ are illustrated following the questions below.

With respect to formula (2) above, if one defines $$a(m,n)=\frac{(-1)^{n-1}}{2^m}\sum\limits_{i=0}^{m-n} \binom{m}{m-n-i}$$ then I believe $$\underset{m\to\infty}{\text{lim}}\left(a(m,n)\right)=(-1)^{n-1}$$ where smaller values of $$n$$ converge much faster than larger values of $$n$$. This explains why $$\eta(s)=\underset{m\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^m\frac{a(m,n)}{n^s}\right)$$ which is equivalent to formula (2) above evaluates much differently than $$\eta(s)=\underset{m\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^m\frac{(-1)^{n-1}}{n^s}\right)$$ which only converges for $$\Re(s)>0$$.

I've reviewed formulas for $$\eta(s)$$ and $$\zeta(s)$$ on Wikipedia and Wolfram MathWorld and haven't found any formulas similar to formulas (2) and (3) above. I believe formulas (2) and (3) evaluate exactly correct when $$s$$ is a non-positive integer and $$|s|\le m-1$$. Exact convergence at non-positive integer values at finite evaluation limits seems to be a characteristic of global (or nearly global) formulas for $$\eta(s)$$ and $$\zeta(s)$$ such as the following which seem to evaluate exactly correct when $$s$$ is a non-positive integer and $$|s|\le K$$.

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

(6) $$\quad\zeta(s)=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{1-2^{1-s}}\sum\limits_{n=0}^K\frac{1}{2^{n+1}}\sum\limits_{k=0}^n\frac{(-1)^k \binom{n}{k}}{(k+1)^s}\right)\,,\quad s\ne 1+\frac{2\,π\,i\,k}{\log(2)}$$

(7) $$\quad\zeta(s)=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{s-1}\sum\limits_{n=0}^K\frac{1}{n+2}\sum\limits_{k=0}^n\frac{(-1)^k \binom{n}{k}}{(k+1)^s}\right)$$

I believe formula (4) above for $$B(s)$$ evaluates exactly correct when $$s$$ is a positive integer and $$s\le m$$ which leads to formula (8) below for $$B_n$$. I've reviewed formulas for $$B_n$$ on Wikipedia and Wolfram MathWorld and the most similar formula I found is illustrated in formula (9) below where $$\mathcal{S}_{n-1}^{(k)}$$ is the Stirling number of the second kind.

(8) $$\quad B_n=\frac{(-1)^n\,n}{2^n\,\left(2^n-1\right)}\sum\limits_{k=1}^n (-1)^{k+1}\,k^{n-1}\sum\limits_{i=0}^{n-k}\binom{n}{n-k-i}\,,\quad n>0$$

(9) $$\quad B_n=\frac{n}{2^n\,\left(2^n-1\right)}\sum\limits_{k=0}^{n-1} (-1)^{k+1}\,k!\,2^{n-k-1}\,\mathcal{S}_{n-1}^{(k)}\,,\quad n>0$$

My interest in the $$B(s)$$ function is motivated in part by the observation that formulas (8) and (9) above are somewhat similar, and it seems to me a proof of formula (8) above for $$B_n$$ (via a proof of the equivalence of these two formulas or some other means) might perhaps lead to a proof of formula (4) above for $$B(s)$$ which would also imply the correctness of formulas (2) and (3) above for $$\eta(s)$$ and $$\zeta(s)$$.

Formulas (8) and (9) above both evaluate correctly for $$n=1$$, and I believe proving the simplified relationship illustrated in formula (10) below is equivalent to proving the correctness of formula (8) above but this is the extent of my progress to date.

(10) $$\quad\sum\limits_{k=1}^n (-1)^{k+1}\,k^{n-1}\sum\limits_{i=0}^{n-k} \binom{n}{n-k-i}=2^n\sum\limits_{k=0}^{n-1}\frac{(-1)^{k+1}}{2^{k+1}}\sum\limits_{i=0}^k (-1)^i\,\binom{k}{i}\,(k-i)^{n-1},\quad n\gt 1$$

Question: Is formula (8) above for $$B_n$$ true, false, or an unprovable statement (an example of Gödel's incompleteness theorems)?

Formula (2) for $$\eta(s)$$ above can be rewritten as illustrated in formula (11) below. I verified formula (11) below evaluates exactly the same as formula (5) for $$\eta(s)$$ above (which is known to be a globally convergent formula) for the first $$100$$ positive integer values of $$K$$, so this is an encouraging result with respect to the conjectured global convergence of formulas (2) and (11).

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

The following two figures illustrate formulas (2) and (3) above for $$\eta(s)$$ and $$\zeta(s)$$ in orange overlaid on the corresponding reference functions in blue where formulas (2) and (3) are both evaluated at $$m=100$$. Note both formulas seem to converge for $$s\in\mathbb R$$.

Figure (3): Illustration of formula (2) for $$\eta(s)$$

Figure (4): Illustration of formula (3) for $$\zeta(s)$$

The following two figures illustrate the real and imaginary parts of formula (4) above for $$B(s)$$ in orange overlaid on the corresponding reference function defined in formula (1) above in blue where formula (4) is evaluated at $$m=100$$ for both figures. The red discrete portion of the first figure below illustrates evaluation of the real part of formula (4) for $$B(s)$$ at positive integer values of $$s$$, and the red discrete portion of the second figure below illustrates evaluation of the imaginary part of formula (4) for $$B(s)$$ at positive and negative integer values of $$s$$. Note the real and imaginary parts of formula (4) both seem to converge for $$s\in\mathbb R$$.

Figure (5): Illustration of real part of formula (4) for $$B(s)$$

Figure (6): Illustration of imaginary part of formula (4) for $$B(s)$$

The following three figures illustrate the absolute value, real part, and imaginary part of formula (3) above for $$\zeta(s)$$ evaluated along the critical line $$s=1/2+i\,t$$ in orange overlaid on the reference function in blue. Formula (3) is evaluated at $$m=100$$ for all three figures. The red discrete portion of each of the three figures below illustrates the evaluation of formula (3) for $$\zeta(s)$$ at the first $$10$$ non-trivial zeta zeros in the upper-half plane. I won't illustrate it here, but I'll note that formulas (2) and (4) for $$\eta(s)$$ and $$B(s)$$ also seem to converge along the critical line $$s=1/2+i\,t$$ in a similar manner.

Figure (7): Illustration of formula (3) for $$|\zeta(1/2+i\,t)|$$

Figure (8): Illustration of formula (3) for $$\Re(\zeta(1/2+i\,t))$$

Figure (9): Illustration of formula (3) for $$\Im(\zeta(1/2+i\,t))$$

• Please use the notation $\eta(s)=\sum\limits_{n=0}^\infty\frac{1}{2^{n+1}}\sum\limits_{k=0}^n\frac{(-1)^k\,\binom{n}{k}}{(k+1)^s}$ and $\lim_{m \to \infty}\frac{1}{2^m}\sum\limits_{n=1}^m\frac{(-1)^{n-1}}{n^s}\sum\limits_{i=1}^{m-n+1}\binom{m}{m-n-i+1}$, yours isn't allowed. How did you come from the former to the latter. I already gave you some proofs of the first. I don't understand why you make your question so long and encumbered instead of just asking if the 2nd limit converges to $\eta(s)$ and mention that numerically it does for a few $s$. Commented Sep 6, 2019 at 12:19
• The main point to prove those kind of formulas is to show it for absolutely convergent series then to do $M$ partial summations (using $\sum_{n =1}^N (-1)^{n+1}= \frac{1+(-1)^{N+1}}{2}$) to get an absolutely convergent series for $\eta(s),\Re(s)>-M+1$. Commented Sep 6, 2019 at 12:20
• @reuns I didn't derive my formula from $\eta(s)=\sum\limits_{n=0}^\infty\frac{1}{2^{n+1}}\sum\limits_{k=0}^n (-1)^k\,\binom{n}{k}\frac{1}{(k+1)^s}$. Formula (2) for $\eta(s)$ in my question here was derived by evaluating formula (1) for $\eta(s)$ in my original question math.stackexchange.com/q/3253496 with $N=0$ and then taking the limit as $m\to\infty$. This approach was motivated by the observation that my original formula seemed to evaluate exactly correct when $s$ is a non-positive integer and $|s|<m$ independent of the value of $N\ge 0$. Commented Sep 6, 2019 at 20:37
• @reuns Assuming $\eta_m(s)=\frac{1}{2^m}\sum\limits_{n=1}^m\frac{(-1)^{n-1}}{n^s}\sum\limits_{i=1}^{m-n+1}\binom{m}{m-n-i+1}$, I'm interested in the exact correctness of $\eta(s)=\eta_m(s)$ when $s$ is a non-positive integer and $|s|<m$ (which is related to the correctness of formula (8) for $B_n$) as well as the pointwise convergence of $\eta(s)=\underset{m\to\infty}{\text{lim}}\eta_m(s)$ for $s\in\mathbb{C}$. Commented Sep 6, 2019 at 21:55
• @reuns Does the approach you described imply anything about the exact correctness of $\eta(s)=\eta_m(s)$ when $s$ is a non-positive integer and $|s|<m$, or does it only imply the pointwise convergence of $\eta(s)=\underset{m\to\infty}{\text{lim}}\eta_m(s)$ for $s\in\mathbb{C}$? Commented Sep 6, 2019 at 21:55