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This question is related to the $f_k(s)$ function defined in (1) below for integer $k>1$ and the two alternate representations defined in (2) and (3) below. I believe formula (2) is valid for all $s$, and formula (3) is valid for $\Re(s)>0$.


Let $q = e^{2\pi i \,r}$, and

$$f_k(s)=\frac{1}{k^s}\sum\limits_{m=1}^k\left(\sum\limits_{r=1}^{k-1}q^{m/k}\right)\zeta\left(s,\frac{m}{k}\right)=\left(k^{1-s}-1\right)\zeta(s)\tag1$$

$$f_k(s)=\sum\limits_{r=1}^{k-1} \text{Li}_s\left(q^{1/k}\right)\qquad\qquad\tag2$$

$$f_k(s)=\sum\limits_{n=1}^N\left(\sum\limits_{r=1}^{k-1}q^{n/k}\right)n^{-s}\,,\quad N\to\infty\land\Re(s)>0\tag3$$


Note that $f_2(s)=\left(2^{1-s}-1\right) \zeta (s)=-\eta(s)$ where $\eta(s)$ is the Dirichlet eta function. Is there a name for the more general class of functions $f_k(s)=\left(k^{1-s}-1\right) \zeta (s)$?


Question (1): Is it true that formula (3) for $f_k(s)$ converges for $\Re(s)>-1$ for odd $k$ when evaluated with an upper limit $N=m\,k+\frac{k-1}{2}$ as the integer $m\to\infty$?


Question (2): Is it true that formula (4) below extends the range of convergence of formula (3) above from $\Re(s)>0$ to $\Re(s)>-1$ as $N\to\infty$?

$f_k(s)=\frac{1}{k}\sum\limits_{m=0}^{k-1}\sum\limits_{n=1}^{N+m}\left(\sum\limits_{r=1}^{k-1}q^{n/k}\right)\,n^{-s}=\sum\limits_{n=1}^N\left(\sum\limits_{r=1}^{k-1}q^{n/k}\right)n^{-s}+\\$ $\frac{1}{k}\sum\limits_{n=1}^{k-1}(k-n)\,\left(\sum\limits_{r=1}^{k-1}q^{(N+n)/k}\right)\,(N+n)^{-s}\,,\,N\to\infty\land\Re(s)>-1?\tag4$


Question (3): The formula for the Dirichlet eta function $\eta(s)=\left(1-2^{1-s}\right)\zeta(s)$ illustrated in formula (5) below is globally convergent. Is there an analogous globally convergent formula for the more general function $f_k(s)=\left(k^{1-s}-1\right)\zeta(s)$?

$$\quad\eta(s)=\sum\limits_{n=0}^N\frac{1}{2^{n+1}}\sum\limits_{k=0}^n\binom{n}{k}\frac{(-1)^k}{(k+1)^s}\,,\quad\,N\to\infty\tag5$$


Formula (3) for $f_k(s)$ exhibits some interesting convergence patterns for $N \bmod k$ as $N\to\infty$ which are illustrated in the four figures below and are the basis of questions (1) and (2) above.


Figure (1) below illustrates formula (3) for $f_2(s)=\left(2^{1-s}-1\right) \zeta(s)$ evaluated at $N=1000\text{ and }1001$ in orange and green respectively overlaid on the corresponding blue reference function.

Illustration of formula (3) for f_2(s)

Figure (1): Illustration of formula (3) for $f_2(s)=\left(2^{1-s}-1\right) \zeta(s)$


Figure (2) below illustrates formula (3) for $f_3(s)=\left(3^{1-s}-1\right) \zeta(s)$ evaluated at $N=1000,1001,\text{ and }1002$ in orange, green, and red respectively overlaid on the corresponding blue reference function. Note for $N=1000$ (orange), $1000\bmod 3=1=\frac{3-1}{2}$.

Illustration of formula (3) for f_3(s)

Figure (2): Illustration of formula (3) for $f_3(s)=\left(3^{1-s}-1\right) \zeta(s)$


Figure (3) below illustrates formula (3) for $f_4(s)=\left(4^{1-s}-1\right) \zeta(s)$ evaluated at $N=1000,1001,1002,\text{ and }1003$ in orange, green, red, and purple respectively overlaid on the corresponding blue reference function.

Illustration of formula (3) for f_4(s)

Figure (3): Illustration of formula (3) for $f_4(s)=\left(4^{1-s}-1\right) \zeta(s)$


Figure (4) below illustrates formula (3) for $f_5(s)=\left(5^{1-s}-1\right) \zeta(s)$ evaluated at $N=1000,1001,1002,1003,\text{ and }$$1004$ in orange, green, red, purple, and brown respectively overlaid on the corresponding blue reference function. Note for $N=1002$ (red curve), $1002\bmod 5=2=\frac{5-1}{2}$.

Illustration of formula (3) for f_5(s)

Figure (4): Illustration of formula (3) for $f_5(s)=\left(5^{1-s}-1\right) \zeta(s)$


To provide a bit more insight, the Dirichlet series for $f_k(s)$ defined in formula (3) above can be written as $f_k(s)=\sum\limits_{n=1}^\infty a_k(n)\,n^{-s}$ where $a_k(n)=\sum\limits_{r=1}^{k-1} q^{\frac{n}{k}}=\sum\limits_{r=1}^{k-1} e^{2 \pi i\, r \frac{n}{k}}$. The following table of examples illustrates $a_k(n)$ is periodic of period $k$.

$$\begin{array}{cc} k & a_k(n) \\ 2 & \{-1,1,-1,1,-1,1,-1,1,-1,1,-1,1\} \\ 3 & \{-1,-1,2,-1,-1,2,-1,-1,2,-1,-1,2\} \\ 4 & \{-1,-1,-1,3,-1,-1,-1,3,-1,-1,-1,3\} \\ \end{array}$$

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  • $\begingroup$ As your post is rather long and complicated, I've made some formatting changes to highlight your primary question so readers can focus on it. (Your other question was just terminology.) I hope the changes are ok with you. $\endgroup$ Commented May 31, 2019 at 3:46
  • $\begingroup$ @Tito Piezas III I like what you did with the early part. I refined the latter part a bit further in an attempt to make thinks a bit clearer. $\endgroup$ Commented May 31, 2019 at 4:18
  • $\begingroup$ anything unclear with my answer ? .. do you see how any zero-mean $q$-periodic dirichlet series can be put is this form (this is quite the main reason why they are entire) $\endgroup$
    – reuns
    Commented May 31, 2019 at 23:26
  • $\begingroup$ @reuns I was stuck on the simple case $q=2$ where $a_n=(-1)^{n-1}$ corresponding to the Dirichlet eta function $\eta(s)=(1-2^{1-s})\,\zeta(s)$, and I still am even after reading your comment on my newer related question. You indicated to go from $f_a(s)=\sum_{n=1}^{\infty}a_n\,n^{-s}$ to $f_b(s)=\sum_{n=1}^{\infty}b_n\left(n^{-s}-(n+1)^{-s}\right)$, $b_n=\sum_{m=1}^n a(m)$. For $a_n=(-1)^{n-1}$, $b_n=n\bmod 2$ which only seems to converge for $Re(s)>0$. Note $\eta(0)=\frac{1}{2}$ whereas $f_b(0)=0$. $\endgroup$ Commented Jun 6, 2019 at 1:07
  • $\begingroup$ @reuns I believe $\eta(s)=\frac{1}{2}+\frac{1}{2}\sum_{n=1}^{\infty }(-1)^{n-1}\left(n^{-s}-(n+1)^{-s}\right)$ converges for $\Re(s)>-1$. $\endgroup$ Commented Jun 6, 2019 at 14:36

1 Answer 1

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$$F(s) = \sum_{n=N}^\infty a_n (n^{-s} - (n+1)^{-s})$$

If $a_n = a_{n+q}$ and $\sum_{n=1}^q a_n = 0$ then $$F(s) = \sum_{n=N}^\infty a_n (s n^{-s-1}+O(s (s+1)n^{-s-2})$$ converges and is analytic for $\Re(s) > -1$.

Example : $$(1-3^{1-s}) \zeta(s) =1+ \sum_{n=2}^\infty a_n (n^{-s} - (n+1)^{-s}), \qquad a_n=a_{n+3},a_1=0,a_2 = 1,a_3 = -1$$

$\scriptstyle n^{-s} - (n+1)^{-s} = \int_n^{n+1} s t^{-s-1}dt=sn^{-s-1}+\int_n^{n+1} s (t^{-s-1}-n^{-s-1})dt=sn^{-s-1}-\int_n^{n+1} s\int_n^t (s+1)u^{-s-2}dudt=sn^{-s-1}+O(s(s+1)n^{-s-2})$

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