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Question: Is it true that for a zero-mean q-periodic Dirichlet series (where $a_n=a_{n+q}$ and $\sum_{n=1}^q a_n=0$) formula (2) below extends the range of convergence of formula (1) below from $\Re(s)>0$ to $\Re(s)>-1$ as $N\to\infty$?

(1) $\quad f(s)=\sum_\limits{n=1}^N a_n\,n^{-s}\,,\quad\Re(s)>0\land N\to\infty$

(2) $\quad f(s)=\frac{1}{q}\sum\limits_{m=0}^{q-1}\sum\limits_{n=1}^{N+m} a_n\,n^{-s}\\$$\qquad\qquad\,\,\,=\sum\limits_{n=1}^N a_n\,n^{-s}+\frac{1}{q}\sum\limits_{n=1}^{q-1}(q-n)\,a_{N+n}\,(N+n)^{-s}\,,\quad\Re(s)>-1?\land N\to\infty$


As the simplest example, the following figures illustrate formulas (1) and (2) above for the Dirichlet eta function $\eta(s)=(1-2^{1-s})\,\zeta(s)$ evaluated with upper limits $N=100$, $N=101$, $N=1000$, and $N=1001$ in orange, green, red, and purple respectively overlaid on the blue reference function. Note formula (2) seems to converge in the interval $-1<x<0$ (see figure (4) below) in a manner analogous to the convergence of formula (1) in the interval $0<x<1$ (see figure (2) below). Note for $\eta(s)$, $a_n=(-1)^{n-1}$ and $q=2$.


Illustration of Formula (1) for eta(s)

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


Illustration of Formula (1) for eta(s)

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


Illustration of Formula (2) for eta(s)

Figure (3): Illustration of Formula (2) for $\eta(s)$


Illustration of Formula (2) for eta(s)

Figure (4): Illustration of Formula (2) for $\eta(s)$

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  • $\begingroup$ No. Why don't you use the method that works math.stackexchange.com/a/3246926/276986 are you stuck in going from $\sum_{n=1}^\infty a_n n^{-s}$ to $\sum_{n=1}^\infty b_n (n^{-s}-(n+1)^{-s})$ ? $b_n = \sum_{m=1}^n a_m$ since $a_n$ is periodic zero-mean $b_n$ is periodic.. $\endgroup$
    – reuns
    Jun 5, 2019 at 20:59
  • $\begingroup$ @reuns Quite some time ago I noticed $\eta(s)=\sum_{k=1}^N\frac{(-1)^{n-1}}{n^s}+\frac{1}{2}\frac{(-1)^N}{(N+1)^s}$ seems to exhibit observational evidence of convergence for $\Re(s)>-1$ as $N\to\infty$, and have been wondering if this is the first step in an iterative approach leading to $\eta(s)=\sum_{n=0}^N\frac{1}{2^{n+1}}\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{(k+1)^s}$ which is globally convergent as $N\to\infty$. $\endgroup$ Jun 5, 2019 at 23:29
  • $\begingroup$ @reuns More recently I noticed $f_k(s)=\left(k^{1-s}-1\right)\zeta(s)$ in my related question evaluated via formula (2) above seems to exhibit observational evidence of convergence for $\Re(s)>-1$ as $N\to\infty$ at least for a few values of $k$ which I've investigated. I evaluated a few other zero-mean q-periodic Dirichlet series via formula (2) above which also seem to exhibit observational evidence of convergence for $\Re(s)>1$ as $N\to\infty$. $\endgroup$ Jun 5, 2019 at 23:30
  • $\begingroup$ @reuns I agree the formula you pointed out perhaps provides more insight, but formula (2) above evaluates almost twice as fast. I've been wondering if it really converges if not in general then at least for a few of the cases I've investigated. For example, can you explain why formula (2) for the Dirichlet eta function doesn't converge for $\Re(s)>-1$ as $N\to\infty$ (see the figures I added to the question above). $\endgroup$ Jun 5, 2019 at 23:31
  • $\begingroup$ $\Re(s)>1$ should have been $\Re(s)>-1$ at the end of my second comment above. $\endgroup$ Jun 6, 2019 at 0:14

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