Convergence of Euler-transformed zeta series

Question

I am trying to prove that the expression

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

converges to an analytic continuation of the alternating zeta function for all $s\in\Bbb C\setminus\{1\}$, following the paper by Sondow.

My strategy is to show that this "Euler transformation" of the original series,

$$(1-2^{1-s})\zeta(s)=\sum_{n=1}^\infty(-1)^{n-1}n^{-s},$$

which is convergent for $\Re s>0$, can be achieved by repeatedly applying the transformation

$$a_1-a_2+a_3-\dots=\frac12a_1+\frac12[(a_1-a_2)-(a_2-a_3)+(a_3-a_4)-\cdots],$$

which after $k$ transformations extends the region of convergence to $\Re s>-k$, so that the limit sequence is convergent everywhere. The general term after $k$ transformations of the alternating zeta function is:

$$(1-2^{1-s})\zeta(s)=\sum_{n=0}^{k-1}\sum_{m=0}^n\frac{(-1)^m}{2^{n+1}}{n\choose m}(m+1)^{-s}+\frac1{2^k}\sum_{n=1}^\infty\sum_{m=0}^k(-1)^{n+m-1}{k\choose m}(m+n)^{-s}$$

Now it follows from the alternating series test that for $k=0$, the original series is convergent for $\Re s>0$, but in order to apply the alternating series test on the transformed series, I need to know that

$$\sum_{m=0}^k(-1)^{m-1}{k\choose m}(m+n)^{-s}\ge0\tag{1}$$ $$\sum_{m=0}^k(-1)^{m-1}{k\choose m}(m+n+1)^{-s}\le\sum_{m=0}^k(-1)^{m-1}{k\choose m}(m+n)^{-s}.\tag{2}$$

Now $(1)$ for $k$ follows from $(2)$ for $k-1$, but verifying $(2)$ is not at all obvious to me. Am I going about this the right way?

Answer 1

Note: According to OPs comment the answer he is interested in is already given in the predecessor question.

In order to provide some additional information here's a different approach following the reference stated in the paper by J. Sondow. In fact this answer addresses only a specific aspect of OPs question.

J. Sondow states in section 2 that any converging series $A$ of complex numbers, written with alternating signs has following representation \begin{align*} A=\sum_{n=1}^{\infty}(-1)^{n-1}a_n=\sum_{j=0}^{k-1}\frac{\Delta^j a_1}{2^{j+1}}+ \sum_{n=1}^{\infty}(-1)^{n-1}\frac{\Delta^k a_n}{2^k},\tag{1} \end{align*} where $\Delta^0 a_n=a_n$ and \begin{align*} \Delta^k a_n=\Delta^{k-1}a_n-\Delta^{k-1}a_{n+1}=\sum_{m=0}^k(-1)^m\binom{k}{m}a_{n+m}, \end{align*} for $k\geq 1$. He continues that in Theory and Application of Infinite Series by Konrad Knopp it is proved, that the sum of the last series in (1) approaches $0$ as $k\rightarrow \infty$, so that \begin{align*} A=\sum_{n=1}^{\infty}(-1)^{n-1}a_n=\sum_{j=0}^{k-1}\frac{\Delta^j a_1}{2^{j+1}}\tag{2} \end{align*} which is Euler's transformation of series.

Proof of Euler's transformation of series: based upon K. Knopp, § $33B$

In order to proof this statement we consider more generally a convergent series $\sum_{k=0}^{\infty}z^{(k)}$ with each of its terms is expressed, in any manner as the sum of an infinite series:

\begin{align*} (A)\quad\begin{cases} z^{(0)}&=a_0^{(0)}+a_1^{(0)}+\cdots+a_n^{(0)}+\cdots\\ z^{(1)}&=a_0^{(1)}+a_1^{(1)}+\cdots+a_n^{(1)}+\cdots\\ &\cdots\\ z^{(k)}&=a_0^{(k)}+a_1^{(k)}+\cdots+a_n^{(k)}+\cdots\\ &\cdots\\ \end{cases} \end{align*}

We shall assume further that the vertical columns in this array themselves constitute convergent series, and denote their sums by $s^{(0)},s^{(1)},\ldots,s^{(n)},\ldots$

Question: Under what conditions may the series $$\sum_{n=0}^{\infty}s^{(n)}$$ formed by these numbers be expected to converge, with $$\sum_{k=0}^{\infty}z^{(k)}=\sum_{n=0}^{\infty}s^{(n)}$$

Note: Knopp uses $z$ to denote Zeilen, the german word for rows and he uses $s$ to denote Spalten, the german word for columns.

Knopp then shows a far-reaching theorem originally proved by A. Markoff:

Markoff's transformation of series:

Let a convergent series $A=\sum_{k=0}^{\infty}z^{(k)}$ be given with each of its terms itself expressed as a convergent series: $$z^{(k)}=a_0^{(k)}+a_1^{(k)}+\cdots+a_n^{(k)}+\cdots\qquad k\geq 0$$ Let the individual columns $\sum_{k=0}^{\infty}a_n^{(k)}$ of the array $(A)$ so formed represent convergent series with sums $s^{(n)}, n\geq 0$, so that the remainders $$r_m^{(k)}=\sum_{n=m}^{\infty}a_n^{(k)}\qquad\qquad m\geq 0$$ of the series in the horizontal rows also constitute a convergent series $$\sum_{k=0}^{\infty}r_m^{(k)}=R_m\qquad\qquad m \text{ fixed}.$$

In order that the sums by vertical columns should form a convergent series $\sum_{n=0}^{\infty}s^{(n)}$, it is necessary and sufficient that $lim_{n\rightarrow \infty}R_m=R$ should exist; and in order that the relation $$\sum_{n=0}^{\infty}s^{(n)}=\sum_{k=0}^{\infty}z^{(k)}$$ should hold as well, it is necessary and sufficient that this limit $R$ should be $0$.

Note: Observe, that in Markoff's theorem no absolute convergence is required, making this theorem far-reaching and flexible.

With the help of Markoff's transformation of series Knopp then proves Euler's transformation of series as follows:

In the array $(A)$ we substitute for $a_n^{(k)}$: \begin{align*} a_n^{(k)}=(-1)^k\left[\frac{1}{2^n}\Delta^na_k-\frac{1}{2^{n+1}}\Delta^{n+1}a_k\right]\tag{3} \end{align*} If we now sum for every $n$, keeping $k$ fixed (i.e. form the sum of the $k^{th}$ horizontal row) we obtain $$z^{(k)}=\sum_{n=0}^{\infty}a_n^{(k)}=(-1)^k\frac{1}{2^0}\Delta^0a_k=(-1)^ka_k\qquad\qquad k\text{ fixed}.$$

Note: We thereby used following property of a series: If a series $A=\sum_{n=0}^{\infty}a_n$ is given with $a_n=x_n-x_{n+1}$ for all $n$ and $\lim_{n\rightarrow\infty}x_n=\xi$, then \begin{align*} A=\sum_{n=0}^{\infty}a_n=x_0-\xi\tag{4} \end{align*}

For $$\lim_{n\rightarrow\infty}\frac{\Delta^na_k}{2^n}= \lim_{n\rightarrow\infty}\frac{\binom{n}{0}a_k-\binom{n}{1}a_{k+1}+-\cdot+(-1)^n\binom{n}{n}a_{n+k}}{2^n}$$ is equal to zero, because $a_k,-a_{k+1},a_{k+2},\ldots$ form a null sequence.

Accordingly (3) gives an expression for the individual terms of the given series $\sum(-1)^ka_k$ in infinite series. Forming the sum of the $n^{th}$ column, we obtain the series $$\sum_{k=0}^{\infty}(-1)^k\left[\frac{1}{2^n}\Delta^na_k-\frac{1}{2^{n+1}}\Delta^{n+1}a_k\right]\qquad\qquad n\text{ fixed}$$ the generic term of this series, as $\Delta^{n+1}a_k=\Delta^na_k-\Delta^na_{k+1}$, can be written in the form $$\frac{(-1)^k}{2^{n+1}}\left[\Delta^na_k+\Delta^na_{k+1}\right] =\frac{1}{2^{n+1}}\left[(-1)^k\Delta^na_k-(-1)^{k+1}\Delta^na_{k+1}\right],$$ so that the series under consideration may again be summed directly, by (4). We obtain $$\sum_{k=0}^{\infty}a_n^{(k)}=\frac{1}{2^{n+1}}\left[\Delta^na_0-\lim_{k\rightarrow\infty}(-1)^k\Delta^na_k\right]\qquad\qquad k\text{ fixed.}$$

Since, however, the numbers $a_k$ form a null sequence, so do the first differences and the $n^{th}$ differences generally, for any fixed $n$. The vertical columns are thus seen to constitute convergent series of sums $$s^{(n)}=\frac{\Delta^na_0}{2^{n+1}}.$$

The validity of Euler's transformation will accordingly be established, when we have shown that $R_m\rightarrow0$. Now the horizontal remainders are seen to have the values $$r_m^{(k)}=(-1)^k\frac{\Delta^ma_k}{2^m,}$$ following precisely the same line of argument as was above for the entire horizontal rows. Thus $$R_m=\frac{1}{2^m}\sum_{k=0}^{\infty}(-1)^k\Delta^ma_k\qquad\qquad m\text{ fixed.}$$ If we write for brevity $(-1)^k(a_k-a_{k+1}+a_{k+2}-+\cdots)=r_k$, this series for $R_m$ may be thought of as obtained by term-by-term addition from the $(m+1)$ series: $$r_0,\binom{m}{1}r_1,\binom{m}{2}r_2,\ldots,\binom{m}{m}r_m.$$ Hence $$R_m=\frac{r_0+\binom{m}{1}r_1+\binom{m}{2}r_2+\cdots+\binom{m}{m}r_m}{2^m}$$ therefore, as $r_m$ is the term of a null sequence, so is $R_m$ and this proves the validity of Euler's transformation with full generality.