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

(1) $quadzeta(s)=frac{1}{1-2^{1-s}}sumlimits_{n=0}^inftyfrac{1}{2^{n+1}}sumlimits_{k=0}^nfrac{(-1)^kbinom{n}{k}}{(k+1)^s},quad sne 1quadtext{(see ref(1) and formula (21) at ref(2))}$

(2) $quadzeta(s)=frac{1}{s-1}sumlimits_{n=0}^inftyfrac{1}{n+1}sumlimits_{k=0}^nfrac{(-1)^kbinom{n}{k}}{(k+1)^{s-1}}qquadqquadqquadtext{(see ref(1) and formula (22) at ref(2))}$

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

Question (1): Is it true that formula (1) converges for $sne 1+ifrac{2,pi,j}{log(2)}$ where $jin mathbb{Z}$ versus $sne 1$? Or is there an argument about zeros and poles cancelling each other out when formula (1) for $zeta(s)$ is evaluated at $s=1+ifrac{2,pi,j}{log(2)}$ where $jin mathbb{Z}$ similar to the argument for the convergence of the right side of the functional equation $zeta(s)=2^s π^{s−1}sinleft(frac{π,s}{2}right),Gamma(1−s),zeta(1−s)$ 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 $sinmathbb{Z}$.

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

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.

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

• 
  

  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 $Im(s)$.