# Questions on two Formulas for $\zeta(s)$

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?

## 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.

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 $$\zeta(s)$$

• take a look at (24) in ref 2 – Will Jagy Dec 10 '18 at 1:01
• 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. – Somos Dec 10 '18 at 1:59
• 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. – Somos Dec 10 '18 at 2:21
• @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. – Steven Clark Dec 10 '18 at 2:32
• @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 $\Re(s)>-1$. – 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 $$Im(s)$$.