Asymptotics of Certain q-Series

TL;DR

This paper derives complete asymptotic expansions for specific q-series as q approaches 1 from below, revealing detailed behavior of these series in the scale of powers of log q.

Contribution

It provides new asymptotic formulas for two classes of q-series involving divisor functions and power sums, extending understanding of their behavior near q=1.

Findings

Asymptotic expansions in powers of log q as q approaches 1

Explicit formulas for series involving divisor functions

Enhanced understanding of q-series behavior near the singularity

Abstract

In this work we study complete asymptotic expansions for the q-series $\sum_{n=1}^{\infty}\frac{1}{n^{b}}q^{n^{a}}$ and $\sum_{n=1}^{\infty}\frac{\sigma_{\alpha}(n)}{n^{b}}q^{n^{a}}$ in the scale function $(\log q)^{n}$ as $q\to1^{-}$, where $a>0,\ q\in(0,1),\,b,\alpha\in\mathbb{C}$ and $\sigma_{\alpha}(n)$ is the divisor function $\sigma_{\alpha}(n)=\sum_{d\vert n}d^{\alpha}$.

Equations

$$\Gamma \left(z\right)={\int }_0^{\infty }{e}^{-x}{x}^{z-1}dx,\Re (z)>0,$$

$$\Gamma \left(z\right)={\int }_1^{\infty }{e}^{-x}{x}^{z-1}dx+\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n}{n!\left(n+z\right)},z\in \mathbb{C}\backslash {\mathbb{N}}_0.$$

$$\Gamma (\sigma +it)=\mathcal{O}\left(e^{-\pi |t|/2}{\left|t\right|}^{\sigma -1/2}\right),t\in \mathbb{R}$$

$$\psi (z)=\frac{{\Gamma }^{{}^{\prime }}(z)}{\Gamma (z)},z\in \mathbb{C}$$

$$\gamma =-\psi (1)\approx \mathrm{0.577216.}$$

$$\zeta (s)=\sum _{n=1}^{\infty }\frac{1}{n^s},\Re (s)>1,$$

$$\zeta (s)=2^s{\pi }^{s-1}\Gamma (1-s)\sin \left(\frac{\pi s}{2}\right)\zeta (1-s).$$

$$\zeta (\sigma +it)=\mathcal{O}\left({\left|t\right|}^{|\alpha |+1/2}\right)$$

$$\zeta (s)=\frac{1}{s-1}+\sum _{n=0}^{\infty }\frac{(-1{)}^n}{n!}{\gamma }_n(s-1{)}^n,$$

$$\log A=\frac{1}{12}-{\zeta }^{{}^{\prime }}(-1).$$

$$\frac{z}{e^z-1}=\sum _{n=0}^{\infty }\frac{B_n}{n!}{z}^n,\left|z\right|<2\pi .$$

$$B_0=1,B_1=-\frac{1}{2},B_{2n-1}=0,n\in \mathbb{N}.$$

$$\zeta (-2n)=0,\zeta (1-2n)=-\frac{B_{2n}}{2n},n\in \mathbb{N}.$$

$${\sigma }_{\alpha }(n)=\sum _{d\mid n}{d}^{\alpha },$$

$$\sum _{n=1}^{\infty }\frac{{\sigma }_{\alpha }(n)}{n^s}=\zeta (s)\zeta (s-\alpha ),\Re (s)>\max \left\{1,\Re (\alpha )+1\right\}$$

$$\sum _{n=1}^{\infty }\frac{d(n)}{n^s}={\zeta }^2(s)\Re (s)>1.$$

$$\prod _{j=1}^k\zeta \left(a_{j}s+b_j\right)=\sum _{n=1}^{\infty }\frac{f_k(n)}{n^s},\Re (s)>\underset{1\le j\le k}{\max }\left\{\frac{1-\Re (b_j)}{a_j}\right\},$$

$$\frac{1}{2\pi i}{\int }_{c-i\infty }^{c+i\infty }\Gamma (s)\prod _{j=1}^k\zeta \left(a{a}_{j}s+b_j+b\right)\frac{ds}{x^s}=\sum _{n=1}^{\infty }\frac{f_k(n)}{n^b}{e}^{-n^{a}x}.$$

$$\sum _{n=1}^{\infty }\frac{f_k(n)}{n^b}{e}^{-n^{a}x}$$

$$-n\ne \frac{1-b-b_j}{a{a}_j},j=1,\dots ,k.$$

$${\sigma }_0=\Re (s_0)>\underset{1\le j\le k}{\max }\left\{\frac{1-\Re (b_j+b)}{a{a}_j}\right\},$$

$$M=\sum _{n=1}^{\infty }\frac{\left|f_k(n)\right|}{n^{a{\sigma }_0+\Re (b)}},s=s_0+1,a=N-1,b=N$$

$$\sum _{n=1}^{\infty }\frac{\left|f_k(N)\right|}{n^{\Re (b)}}{e}^{-n^{a}x}<\infty ,a,x>0,b\in \mathbb{C}.$$

$$\frac{1}{2\pi i}{\int }_{c-i\infty }^{c+i\infty }\Gamma (s)\frac{ds}{x^s}=e^{-x}$$

$$g(s)=\Gamma (s)\prod _{j=1}^k\zeta \left(a{a}_{j}s+b_j+b\right)x^{-s},$$

$$\frac{1}{2\pi i}{\int }_{c-i\infty }^{c+i\infty }g(s)ds=\frac{1}{2\pi i}{\int }_{c-i\infty }^{c+i\infty }\Gamma (s)\left(\sum _{n=1}^{\infty }\frac{f_k(n)}{n^{as+b}}\right)\frac{ds}{x^s}$$

$$=\sum _{n=1}^{\infty }\frac{f_k(n)}{n^b}\frac{1}{2\pi i}{\int }_{c-i\infty }^{c+i\infty }\Gamma (s)\frac{ds}{{\left(n^{a}x\right)}^s}=\sum _{n=1}^{\infty }\frac{f_k(n)}{n^b}{e}^{-n^{a}x},$$

$$s=\frac{1-b-b_j}{a{a}_j},j=1,\dots ,k$$

$$N>\underset{1\le j\le k}{\max }\left\{\frac{1+|b+b_j|}{a{a}_j}\right\}+1,M>\underset{1\le j\le k}{\max }\left\{\frac{1+|b+b_j|}{a{a}_j}\right\},$$

$$c-iM,c+iM,-N-\frac{1}{2}+iM,-N-\frac{1}{2}-iM.$$

$${\int }_{\mathcal{R}(M,N)}\frac{g(s)ds}{2\pi i}=\sum _j\text{Residue}\left\{g(s),s=\frac{1-b-b_j}{a{a}_j}\right\}+\sum _n\text{Residue}\left\{g(s),s=-n\right\},$$

Full text

The work is supported by the National Natural Science Foundation of China grants No. 11371294 and No. 11771355.

1. Preliminaries

In this work we study complete asymptotic expansions for the q-series $\sum_{n=1}^{\infty}\frac{1}{n^{b}}q^{n^{a}}$ and $\sum_{n=1}^{\infty}\frac{\sigma_{\alpha}(n)}{n^{b}}q^{n^{a}}$ in the scale function $(\log q)^{n}$ as $q\to 1^{-}$, where $a>0,\ q\in(0,1),\,b,\alpha\in\mathbb{C}$ and $\sigma_{\alpha}(n)$ is the divisor function $\sigma_{\alpha}(n)=\sum_{d|n}d^{\alpha}$. Unlike methods used [3, 4], our method does not apply Fourier transform or the modular properties, it can not give $\sum_{n=1}^{\infty}\frac{1}{n^{b}}q^{n^{a}}$ a complete asymptotic expansion in exponential scales when $a=2$ and $b$ is an even integer. However, this shortcoming can be overcome by applying the functional equations for the corresponding zeta functions which are equivalent to the symmetry $x\to 1/x$.

The Euler gamma function is defined by

[TABLE]

and its analytic continuation is given by

[TABLE]

Let $a,b\in\mathbb{R}$ and $a<b$, it is known that [1, 5, 6]

[TABLE]

as $t\to\pm\infty$, uniformly with respect to $\sigma\in[a,b]$. The digamma function is defined by

[TABLE]

and the Euler’s constant is

[TABLE]

The Riemann zeta function $\zeta\left(s\right)$ is defined by

[TABLE]

then its analytic continuation, which is also denoted as $\zeta(s)$, is an meromorphic function that has a simple pole at $1$ with residue $1$. The meromorphic function $\zeta(s)$ satisfies the functional equation [1, 2, 5, 6]

[TABLE]

For $\alpha,\beta\in\mathbb{R}$ and $\alpha\leq\sigma\leq\beta$, it is known that [6]

[TABLE]

as $t\to\pm\infty$, uniformly with respect to $\sigma$. The Stieltjes constants $\gamma_{n}$ are the coefficients in the Laurent expansion,

[TABLE]

where $\gamma_{0}=\gamma$ and $\gamma_{1}\approx-0.0728158$. Moreover, the Glaisher’s constant $A\approx 1.28243$ is defined as

[TABLE]

The Bernoulli numbers $B_{n}$ are defined by

[TABLE]

Then

[TABLE]

By (1.7) we get

[TABLE]

The function $\sigma_{\alpha}(n)$ for $\alpha\in\mathbb{C}$ is defined as the sum of the $\alpha$-th powers of the positive divisors of $n$, [2, 5]

[TABLE]

where $d|n$ stands for "$d$ divides $n$". We also use the notations $d(n)=\sigma_{0}$ and $\sigma(n)=\sigma_{1}(n).$ It is known that [2, 5]

[TABLE]

and

[TABLE]

2. Main Results

Theorem 1**.**

Given a positive integer $k$, let $a_{j}\in\mathbb{N},\ b_{j}\in\mathbb{C}$ for all $j$ satisfying $1\leq j\leq k$.

If

[TABLE]

where $\zeta(s)$ is the Riemann zeta function, then for all $x,a>0$, $b\in\mathbb{C}$ and $c>0$ satisfying $c>\max_{1\leq j\leq k}\left\{\frac{1-\Re(b_{j}+b)}{aa_{j}}\right\}$ we have

[TABLE]

Furthermore,

[TABLE]

as $x\to 0^{+}$, where the first sum is over all the distinct pairs $a_{j}\,b_{j}\ j=1,\dots,k$ while the last sum is over all nonnegative integers $n$ such that

[TABLE]

Proof.

For $\Re(s)>\max_{1\leq j\leq k}\left\{\frac{1-\Re(b_{j}+b)}{aa_{j}}\right\}$, since each factor of $\prod_{j=1}^{k}\zeta\left(aa_{j}s+b_{j}+b\right)$ is an absolute convergent Dirichlet series, then the product itself is also an absolute convergent Dirichlet series. Let $s_{0}$ be any complex number satisfying

[TABLE]

then by the theory of Dirichlet series we know the partial sums $\sum_{n\leq x}\frac{f_{k}(n)}{n^{as_{0}+b}}$ are absolutely and uniformly bounded for all $x>1$. Let $N$ be a large positive integer and

[TABLE]

in Lemma 2 of section 11.6 in [2] to get $\left|f_{k}(N)\right|\leq 4MN^{\sigma_{0}}.$ Hence,

[TABLE]

By the inverse Mellin transform of (1.1) we get

[TABLE]

for all $x,c>0$.

Let

[TABLE]

then for any positive $c$ satisfying $c>\max_{1\leq j\leq k}\left\{\frac{1-\Re(b_{j}+b)}{aa_{j}}\right\}$, by (2.6) we get

[TABLE]

where we have applied (2.5) and the Fubini’s theorem to exchange the order of summation and integration.

Since $\zeta(s)$ has a simple pole at $s=1$ and $\Gamma(s)$ has simple poles at all non-positive integers, then all the possible poles of the meromorphic function $g(s)$ are

[TABLE]

and all non-positive integers. Let $N\in\mathbb{N}$ and $M\in\mathbb{R}$ such that

[TABLE]

we integrate $g(s)$ over the rectangular contour $\mathcal{R}(M,N)$ with vertices,

[TABLE]

Then by Cauchy’s theorem we have

[TABLE]

where the first sum is over all the distinct pairs from $a_{j}\,b_{j},\ j=1,\dots,k$ whereas the last sum is over all $n$ satisfying $0\leq n\leq N$ and (2.4).

On the other hand, we also have

[TABLE]

Fix $N$ and $x$, by (1.3) and (1.8), since the integrands of the last two integrals have the estimate

[TABLE]

where $\epsilon$ is an arbitrary positive number such that $0<\epsilon<\frac{\pi}{2}$, then the last two integrals have limit [math] as $M\to\infty$. Then by taking limit $M\to\infty$ in (2.9) and (2.8) we get

[TABLE]

where the summations are the same as in (2.8).

Since

[TABLE]

then again by (1.3) and (1.8) we get

[TABLE]

as $x\to 0$. Then by (2.10) and (2.11) we get

[TABLE]

as $x\to 0$, where the first sum is over all the distinct pairs from $a_{j}\,b_{j},\ j=1,\dots,k$ while the last sum is over all nonnegative integers $n$ satisfying (2.4). Finally, (2.2) is obtained by combining (2.7) and (2.12). ∎

Corollary 2**.**

Let $a,x>0,\ b\in\mathbb{C}$. If $b\neq 1+an,\quad n\in\mathbb{N}\cup\left\{0\right\}$, then

[TABLE]

as $x\to 0$.

If there exists a $n_{0}\in\mathbb{N}\cup\left\{0\right\}$ such that $b=1+an_{0}$, then

[TABLE]

as $x\to 0$.**

Proof.

When $b\neq 1+an,\quad n\in\mathbb{N}\cup\left\{0\right\},$ the integrand $\frac{\Gamma(s)\zeta(as+b)}{x^{s}}$ is meromorphic and has the following simple poles

[TABLE]

with residues

[TABLE]

Then (2.13) is obtained by applying Theorem 1.

When $b=1+an_{0}$ for some nonnegative integer $n_{0}$, then

[TABLE]

has a double pole at $-n_{0}$ with residue

[TABLE]

all the other nonpositive integers are simple poles with residues,

[TABLE]

Then by Theorem 1 we have

[TABLE]

as $x\to 0$. ∎

Example 3**.**

When $a=2,\ b=0$ we have

[TABLE]

as $x\to 0$, which means the error term is better than any $x^{n}$. When $a=2,\ b=1$, the double pole happens at $n_{0}=0$, then

[TABLE]

as $x\to 0$. When $a=2,\ b=-1$, then $2n+1=-1$ has no nonnegative integer solutions. Thus,

[TABLE]

or

[TABLE]

as $x\to 0$.

Corollary 4**.**

For all $x,a>0$, $b\in\mathbb{C}$ and $c>\frac{1-\Re(b)}{a}$ we have

[TABLE]

Furthermore, if $an\neq b-1$ for all nonnegative integers $n$, then

[TABLE]

as $x\to 0^{+}$. **

If $am=b-1$ for certain nonnegative $m$, then

[TABLE]

as $x\to 0^{+}$.**

Proof.

When $an\neq b-1$ for all nonnegative integers $n$, then the meromorphic function $\Gamma(s)\zeta^{2}(as+b)x^{-s}$ has a double pole at $s=(1-b)/a$ with residue

[TABLE]

and simple poles at all nonpositive integers $n\in\mathbb{N}_{0}$ with residue

[TABLE]

Then by Theorem 1 we get

[TABLE]

as $x\to 0$.

When $\alpha=0,\ am=b-1$ for certain nonnegative integer $m$, then the meromorphic function $\Gamma(s)\zeta^{2}(as+b)x^{-s}$ has a triple pole at $s=-m$ with residue

[TABLE]

It has simple poles at all other nonpositive integers with residue

[TABLE]

Then by Theorem 1 we get

[TABLE]

as $x\to 0^{+}$. ∎

Example 5**.**

Let $a=2,\ b=2$, then by (2.22) to get

[TABLE]

as $x\to 0^{+}$, the remainder here is better than any $x^{n}$. Let $a=2,\ b=1$, then by (2.23) to get

[TABLE]

as $x\to 0^{+}$.

Corollary 6**.**

Let $\alpha\in\mathbb{C}$ and $\alpha\neq 0$, then for all $x,a>0$, $b\in\mathbb{C}$ and $c>\frac{\max_{1\leq j\leq k}\left\{1-\Re(b),1-\Re(b-\alpha)\right\}}{a}$ we have

[TABLE]

Furthermore, if $an\neq b-1$ and $an\neq b-1-\alpha$ for all nonnegative integers $n\in\mathbb{N}_{0}$, then

[TABLE]

as $x\to 0^{+}$.

If $am=b-1$ for certain nonnegative integer $m$ and $an\neq b-1-\alpha$ for all nonnegative integers $n\in\mathbb{N}_{0}$, then

[TABLE]

as $x\to 0^{+}$.

If $an\neq b-1$ for all nonnegative integers $n\in\mathbb{N}_{0}$ and $am=b-1-\alpha$ for certain $m\in\mathbb{N}_{0}$, then

[TABLE]

as $x\to 0^{+}$.

If $am=b-1$ and $\alpha=a(m-\ell)$ for certain nonnegative integers $m,\ell\in\mathbb{N}_{0}$ with $m\neq\ell$, then

[TABLE]

as $x\to 0^{+}$.

Proof.

When $\alpha\neq 0$, $an\neq b-1$ and $an\neq b-1-\alpha$ for all nonnegative integers $n\in\mathbb{N}_{0}$, the meromorphic function $\Gamma(s)\zeta(as+b)\zeta(as+b-\alpha)x^{-s}$ has simple poles at

[TABLE]

with residues

[TABLE]

[TABLE]

and

[TABLE]

for $n\in\mathbb{N}_{0}$. Then by Theorem 1 we get

[TABLE]

as $x\to 0^{+}$.

When $am=b-1$ for certain nonnegative integer $m$ and $an\neq b-1-\alpha$ for all nonnegative integers $n\in\mathbb{N}_{0}$, then the meromorphic function $\Gamma(s)\zeta(as+b)\zeta(as+b-\alpha)x^{-s}$ has a double pole at $s=-m$ with residue

[TABLE]

and a simple pole at $s=-m+\frac{\alpha}{a}$ with residue

[TABLE]

and simple poles at all nonpositive integers other than $-m$ with residues

[TABLE]

Hence,

[TABLE]

as $x\to 0^{+}$.

When $an\neq b-1$ for all nonnegative integers $n\in\mathbb{N}_{0}$ and $am=b-1-\alpha$ for certain $m\in\mathbb{N}_{0}$, then the meromorphic function $\Gamma(s)\zeta(as+b)\zeta(as+b-\alpha)x^{-s}$ has a double simple pole $s=-m$ with residue

[TABLE]

and a simple pole $s=-m-\frac{\alpha}{a}$ with residue

[TABLE]

and simple poles at all nonpositive integers other than $-m$ with residue

[TABLE]

Thus,

[TABLE]

as $x\to 0^{+}$.

When $am=b-1$ and $\alpha=a(m-\ell)$ for certain nonnegative integers $m,\ell\in\mathbb{N}_{0}$ with $m\neq\ell$, then the meromorphic function $\Gamma(s)\zeta(as+b)\zeta(as+b-\alpha)x^{-s}$ has two double poles at $s=-m$ and $s=-\ell$ with residues

[TABLE]

and

[TABLE]

respectively. It has simple poles at all other nonpositive integers other than $-m,\,-\ell$ with residues

[TABLE]

Then,

[TABLE]

as $x\to 0^{+}$. ∎

Example 7**.**

When $a=2,\ \alpha=1,\ b=\frac{1}{2}$, by (2.30) we get

[TABLE]

as $x\to 0^{+}$. When $a=2,\ \alpha=1,\ b=1$, then $m=0$ in (2.31). Then

[TABLE]

as $x\to 0^{+}$, it implies that the difference between two sides of the above formula is smaller than any $x^{n}$. Let $a=2,\ b=1,\ \alpha=-2$ in (2.41), then $m=0,\ \ell=1$. Then,

[TABLE]

as $x\to 0^{+}$.