This question is related to the following two formulas for $\zeta(s)$.

(1) $\quad\zeta(s)=\frac{1}{1-2^{1-s}}\sum\limits_{n=0}^\infty\frac{1}{2^{n+1}}\sum\limits_{k=0}^n\frac{(-1)^k\binom{n}{k}}{(k+1)^s},\quad s\ne 1\quad\text{(see ref(1) and formula (21) at ref(2))}$

(2) $\quad\zeta(s)=\frac{1}{s-1}\sum\limits_{n=0}^\infty\frac{1}{n+1}\sum\limits_{k=0}^n\frac{(-1)^k\binom{n}{k}}{(k+1)^{s-1}}\qquad\qquad\qquad\text{(see ref(1) and formula (22) at ref(2))}$

Formula (1) above is claimed to converge for $s\ne 1$ at ref(2), but note that $\frac{1}{1-2^{1-s}}$ exhibits a complex infinity at $s=1+i\frac{2\,\pi\,j}{\log(2)}$ where $j\in \mathbb{Z}$ which seems consistent with the convergence claim at ref(1).

Question (1): Is it true that formula (1) converges for $s\ne 1+i\frac{2\,\pi\,j}{\log(2)}$ where $j\in \mathbb{Z}$ versus $s\ne 1$? Or is there an argument about zeros and poles cancelling each other out when formula (1) for $\zeta(s)$ is evaluated at $s=1+i\frac{2\,\pi\,j}{\log(2)}$ where $j\in \mathbb{Z}$ similar to the argument for the convergence of the right side of the functional equation $\zeta(s)=2^s π^{s−1}\sin\left(\frac{π\,s}{2}\right)\,\Gamma(1−s)\,\zeta(1−s)$ at positive integer values of s (e.g. see Using the functional equation of the Zeta function to compute positive integer values)?

Since originally posting question (1) above, I discovered the following Wikipedia article which I believe provides some insight.

Wikipedia Article: Landau's problem with $\zeta(s)=\frac{\eta(s)}{0}$ and solutions

Formula (2) above is claimed to be globally convergent, but seems to exhibit a significant divergence (see Figure (1) below).

Question (2): Is there an error in formula (2), or is there a conditional convergence requirement associated with formula (2) when the outer series is evaluated for a finite number of terms?

ref(1): Wikipedia Article: Riemann zeta function, Representations, Globally convergent series

ref(2): Sondow, Jonathan and Weisstein, Eric W. "Riemann Zeta Function." From MathWorld--A Wolfram Web Resource.

12/10/2018 Update:

I'm now wondering if formula (2) for $\zeta(s)$ is perhaps only valid for $s\in\mathbb{Z}$.

The following plot illustrates formula (2) for $\zeta(s)$ evaluated for the first $100$ terms.

Illustration of Formula (2) for zeta(s)

Figure (1): Illustration of Formula (2) for $\zeta(s)$

The following discrete plot illustrates formula (2) for $\zeta(s)$ minus $\zeta(s)$ where formula (2) is evaluated for the first $100$ terms in blue and the first $1000$ terms in orange.

Discrete Plot of Formula (2) for zeta(s)

Figure (2): Discrete Plot of Formula (2) for $\zeta(s)$ minus $\zeta(s)$

  • $\begingroup$ take a look at (24) in ref 2 $\endgroup$ – Will Jagy Dec 10 '18 at 1:01
  • $\begingroup$ You are correct to be skeptical of equation (1). for the reason you gave. Division by zero is never a good idea. Never mind what the infinite sum evaluates to. $\endgroup$ – Somos Dec 10 '18 at 1:59
  • $\begingroup$ About equation (2), ref(2) states "slowly convergent" but it is even worse. Lots of loss of accuracy with cancellation.of sum of an alternating series. For $s=0$, only the first term $n=0$ is nonzero, so it converges easily there. $\endgroup$ – Somos Dec 10 '18 at 2:21
  • $\begingroup$ @WillJagy Formula (24) in ref(2) only seems to converge for real s, whereas formulas that converge along the critical line are my primary interest. Based on formula (24) I thought perhaps there was a missing plus sign in formula (22), so I tried inserting one but it didn't seem to fix the problem. $\endgroup$ – Steven Clark Dec 10 '18 at 2:32
  • $\begingroup$ @Somos Formula (2) seems to perhaps be a generalization of the two formulas defined at en.wikipedia.org/wiki/Riemann_zeta_function#Dirichlet_series which both seem to converge for their stated conditions which are $\Re(s)>0$ and $\Re(s)>-1$. $\endgroup$ – Steven Clark Dec 10 '18 at 2:38
  • Looking at the coefficients of $x_m$ in $$\sum_{k=0}^K 2^{-k-1}\sum_{m=0}^k {k \choose m} x^m = \sum_{k=0} 2^{-k-1}(1+x)^k = \frac{1-2^{-1-K}(1+x)^K}{1-x}$$

    as $K \to \infty$ they converge to $1$ boundedly and locally uniformly,

    so we find that if $\sum_{n=1}^\infty |a_n| < \infty $ then

    $$\sum_{n=1}^\infty a_n = \sum_{k=0}^\infty 2^{-k-1} \sum_{m=0}^k {k \choose m} a_{m+1}$$

  • With $b_m = (-1)^m a_{m+1}$ then $\sum_{m=0}^k {k \choose m} a_{m+1} = \Delta^k b_m$ is the $k$-th forward difference operator

  • Summing by parts $l$ times $(1-2^{1-s}) \zeta(s)= \sum_{n=1}^\infty (-1)^{n+1} n^{-s}$, since $\sum_{n=1}^N (-1)^{N+1} = \frac{1+(-1)^{N+1}}{2}$ and $\Delta^k [(-1)^{n+1}n^{-s}] = O(n^{-s-k})$ we obtain that

    $$(1-2^{1-s}) \zeta(s) = \sum_{r=0}^{l-1} 2^{-r-1} \sum_{m=0}^r {r \choose m} (-1)^{m} (m+1)^{-s}\\ +2^{-l-1}\sum_{n=1}^\infty (-1)^{n+1}\sum_{m=0}^l {l \choose m} (-1)^{m} (n+m)^{-s}$$

    converges absolutely for $\Re(s) > -l+1$.

  • Letting $a_n = \sum_{m=0}^l {l \choose m} (-1)^{n+m+1} (n+m)^{-s}$ so that $$\sum_{m=0}^k {k \choose m} a_{m+1} = \sum_{m=0}^{l+k} {l+k \choose m} (-1)^{n+m+1} (n+m)^{-s}$$ (forward difference operator $\Delta^{l+k}= \Delta^k \Delta^l$)

    we obtain the result

    $$(1-2^{1-s}) \zeta(s) = \sum_{r=0}^\infty 2^{-r-1} \sum_{m=0}^r {r \choose m} (-1)^{m} (m+1)^{-s}$$

    which is valid for every $s$.

Estimating the rate of convergence isn't obvious, it depends on $Im(s)$.


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