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Consider the following two formulas for $\zeta(s)$

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

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

which I believe are both globally convergent (see Globally convergent series).

I believe the formula

$$\zeta(s)=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{\left(1-2^{1-s}\right)} \frac{1}{2^{K+1}}\sum\limits_{n=0}^K \frac{(-1)^n}{(n+1)^s} \sum\limits_{k=0}^{K-n} \binom{K+1}{K-n-k}\right)\tag{3}$$

is exactly equivalent to formula (1) above for non-negative integer values of $K$ and all $s\in\mathbb{C}$.

Note formula (3) above is more efficient than formula (1) above as it moves the exponentiation operation from the inner sum to the outer sum.

Question: Can the exponentiation operation in formula (2) above also be moved from the inner sum to the outer sum analogous to formula (3) above?

I'll note that I derived formula (3) above independent of formula (1) above and subsequently discovered the two formulas seem to be exactly equivalent, but don't yet have a formal proof as no one has answered my related questions on Math StackExchange such as my question on the Bernoulli number $B_n$ and my question on the Dirichlet eta function $\eta(s)$. A formal proof of the equivalence of formulas (1) and (3) above is still of interest to me mainly because it would resolve my curiosity of whether this is an example of an unprovable statement (an example of Gödel's incompleteness theorems).

Formulas (1) to (3) above seem to evaluate exactly correctly for all non-positive integers $|s|\le K$, but this requires evaluating one more value of $n$ in formula (2) above versus formulas (1) and (3) above which is why formula (2) above was specified with the upper evaluation limit $K+1$ whereas formulas (1) and (3) were both specified with the upper evaluation limit $K$.

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  • $\begingroup$ I believe that equation (3) could be simplified further into the also globally convergent series: $$\eta \left( s \right) =\lim _{K\rightarrow \infty } \left( {-2}^{-K-1} \sum _{n=1}^{K}{\frac { \left( -1 \right) ^{n}}{n ^{s}}} {\it \cdot JacobiP} \left( K+1-n,n,-K-1,3 \right)\right)$$ Where $\eta(s)$ is the Dirichlet eta-function. $\endgroup$
    – Agno
    Commented Jun 21, 2022 at 10:37
  • $\begingroup$ @Agno Mathematica likes to simplify it as $\eta(s)=\underset{K\to\infty}{\text{lim}}\left(2^{-K-1} \sum\limits_{n=1}^{K+1} (-1)^{n-1}\, n^{-s} \binom{K+1}{K-n+1} \, _2F_1(1,-K+n-1;n+1;-1)\right)$. I actually derived the formula for $\eta(s)$ first and then modified it for $\zeta(s)$. I tried your formula and it seems to be equivalent after correcting the upper evaluation limit from $K$ to $K+1$ in the sum over $n$, so thanks, I wasn't aware of your simplification. $\endgroup$
    – Steven Clark
    Commented Jun 21, 2022 at 14:42

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