1 Introduction

For any integer \( r\ge 1 \), the multiple zeta function (Euler–Riemann–Zagier type) of depth r is defined by

$$\begin{aligned} \zeta _r(s_1,s_2,\ldots ,s_r)=\sum _{m_1=1}^{\infty }\sum _{m_2=1}^{\infty }\cdots \sum _{m_r=1}^{\infty } \frac{1}{m_1^{s_1}(m_1+m_2)^{s_2}\cdots (m_1+m_2+\cdots +m_r)^{s_r}}\,,\nonumber \\ \end{aligned}$$
(1)

where \( s_i\,(1\le i\le r) \) are complex variables. Throughout the article, we denote \(\Re (s_i)=\sigma _i\). It is well known (see Theorem 3, [4]) that the series (1) is absolutely convergent in the region \( \{(s_1,s_2,\ldots ,s_r)\in {\mathbb {C}}^r:\sigma _r+\sigma _{r-1}+\cdots +\sigma _{r-i}> i+1\,\,\text {for}\,\,0\le i\le r-1\} \). For \( r=1 \), it is nothing but the Riemann zeta function. So multiple zeta function of depth r is multi-variable generalization of Riemann zeta function. Zhao [5] and Akiyama et al. [1] independently have shown that \( \zeta _r(s_1,s_2,\ldots ,s_r) \) can be extended meromorphically to the whole \( {\mathbb {C}}^r \).

One can consider the generalization of the series defined in (1) in the following manner

$$\begin{aligned} \Phi _r((s_j);(a_j))=\sum _{m_1=1}^{\infty }\sum _{m_2=1}^{\infty }\cdots \sum _{m_r=1}^{\infty } \frac{a_1(m_1)a_2(m_2)\cdots a_r(m_r)}{m_1^{s_1}(m_1+m_2)^{s_2}\cdots (m_1+m_2+\cdots +m_r)^{s_r}}\,, \end{aligned}$$
(2)

where \( a_j\,(1\le j\le r)\) are arithmetic functions. For each j, if \( a_j(m)=1,\,\forall \, m\in {\mathbb {N}} \), then \( \Phi _r((s_j);(a_j))= \zeta _r(s_1,s_2,\ldots ,s_r)\).

The first question that one can ask, is to find the region of absolute convergence of the multiple Dirichlet series defined in (2). In this context we have the following result.

Theorem 1

For each \( j\,(1\le j\le r)\), let \(\varphi _{j}(s)=\sum _{m=1}^{\infty }\frac{a_j(m)}{m^s}\) be absolutely convergent for \(\Re (s)>\alpha _j>0\). Then the multiple Dirichlet series defined in (2) is absolutely convergent in the region \( U_r:=\{(s_1,s_2,\ldots ,s_r)\in {\mathbb {C}}^r:\sigma _r+\sigma _{r-1}+\cdots +\sigma _{r-i}>\alpha _r+\alpha _{r-1}+\cdots +\alpha _{r-i}\,\,\text {for}\,\,0\le i\le r-1\}. \)

The series defined in (2) is already considered by Matsumoto and Tanigawa in [3], where they have mentioned that this series is absolutely convergent in the trivial region \( \{(s_1,s_2,\ldots ,s_r)\in {\mathbb {C}}^r:\sigma _i>\alpha _i\,\,\text {for}\,\,1\le i\le r\} \). Since their primary goal was to study the meromorphic continuation, so there is no need to start with exact region of absolute convergence.

Remark 1

In Proposition 3.1 of [2], Matsumoto et al. have given a region of absolute convergence for certain double Dirichlet series that region is equal to \( U_r\) for \( r=2 \) in the Theorem 1.

In the following theorem we give the necessary and sufficient conditions for the series (2) to converge absolutely under certain conditions on the arithmetic functions.

Theorem 2

For each \( j\,(1\le j\le r) \), let the arithmetic function \( a_j(m) \) and the positive real number \( \alpha _j \) satisfy the following conditions \( \bullet \) \( \sum _{m=1}^{\infty }\frac{a_j(m)}{m^s} \) has abscissa of absolute convergence \( \alpha _j, \) \( \bullet \) \( \sum _{m\le t}|a_j(m)|\gg t^{\alpha _j} \) for every \( t\ge 1. \) Then we have that the series defined in (2) is absolutely convergent at \((s_1,s_2,\ldots ,s_r)\in {\mathbb {C}}^r\) if and only if \(\sigma _r+\sigma _{r-1}+\cdots +\sigma _{r-i}>\alpha _r+\alpha _{r-1}+\cdots +\alpha _{r-i}\) for \(0\le i\le r-1\).

As an application of Theorem 2, we have the following result.

Corollary 1

The region of absolute convergence of the series defined in (1) is

$$\begin{aligned} \{(s_1,s_2,\ldots ,s_r)\in {\mathbb {C}}^r:\sigma _r+\sigma _{r-1}+\cdots +\sigma _{r-i}> i+1\,\,\text {for}\,\,0\le i\le r-1\}. \end{aligned}$$

Remark 2

In Proposition 2.1 of [6], Zhao et al. have derived the necessary and sufficient conditions for absolute convergence of certain generalized multiple zeta functions. Corollary 1 is also followed from this proposition.

It is well known that if the Dirichlet series \( \sum _{m=1}^{\infty }\frac{1}{m^s}\) converges at \( s=s_{0} \), then this series converges for all s such that \( \Re (s)>\Re (s_0)\). In the following theorem we prove an analogues result for general multiple Dirichlet series in case of absolute convergence.

Theorem 3

For the arithmetic functions \( a_j(m)\,(1\le j\le r) \), let the series defined in (2) be absolutely convergent at \( (s_1^{\prime },s_2^{\prime },\ldots ,s_r^{\prime })\in {\mathbb {C}}^r\). Then the series converges absolutely at each \( (s_1,s_2,\ldots ,s_r)\in {\mathbb {C}}^r \) such that \(\sigma _r+\sigma _{r-1}+\cdots +\sigma _{r-i}\ge \sigma _r^{\prime }+\sigma _{r-1}^{\prime }+\cdots +\sigma _{r-i}^{\prime }\) for \(0\le i\le r-1\), where \( \Re (s_i^{\prime })=\sigma _i^{\prime } \) for \(1\le i\le r\).

2 Preliminaries

In this section we give some lemmas which are necessary ingredients to prove Theorem 2.

Lemma 1

For \(n\ge 1\) and \( \sigma \in {\mathbb {R}} \), we have

\( \sum _{m=1}^{\infty }\frac{|a(m)|}{m^{\sigma }}<\infty \) if and only if \( \sum _{m=1}^{\infty }\frac{|a(m)|}{(n+m)^{\sigma }}<\infty \) .

Proof

The lemma easily follows from the fact that

$$\begin{aligned} \frac{1}{(n+m)^{\sigma }}\ll \frac{1}{m^{\sigma }}\ll \frac{1}{(n+m)^{\sigma }}\,\,\text {for each}\,\,m\in {\mathbb {N}}\,\,\text {and for any fixed}\,\,n\ge 1. \end{aligned}$$

\(\square \)

Lemma 2

For the positive real number \( \alpha \), let the arithmeitc function a(m) satisfies \( \sum _{m\le t}|a(m)|\gg t^{\alpha } \) for every \( t\ge 1 \). Then \( \sum _{m=1}^{\infty }\frac{|a(m)|}{m^\alpha } \) diverges.

Proof

Using Abel summation formula, we have

$$\begin{aligned} \sum _{m\le t}\frac{|a(m)|}{m^\alpha }=\left( \sum _{m\le t}|a(m)|\right) \frac{1}{t^\alpha }+\alpha \int _{1}^{t}\left( \sum _{m\le x}|a(m)|\right) \frac{1}{x^{\alpha +1}}dx. \end{aligned}$$

This implies that \( \sum _{m\le t}\frac{|a(m)|}{m^\alpha }\gg \alpha \log t \) and hence \( \sum _{m=1}^{\infty }\frac{|a(m)|}{m^\alpha } \) diverges. \(\square \)

Lemma 3

Let the arithmetic function a(m) and positive real number \( \alpha \) satisfy the following conditions

  • \( \sum _{m=1}^{\infty }\frac{a(m)}{m^s} \) has abscissa of absolute convergence \( \alpha , \)

  • \( \sum _{m\le t}|a(m)|\gg t^{\alpha } \) for every \( t\ge 1. \) Then for \( \sigma >\alpha \) and \( n\ge 1 \), we have \( \sum _{m=1}^{\infty }\frac{|a(m)|}{(n+m)^{\sigma }}\gg _{\sigma }\frac{1}{n^{\sigma -\alpha }}\).

Proof

Using Abel summation formula, we have

$$\begin{aligned} \sum _{m\le x}\frac{|a(m)|}{(n+m)^{\sigma }}=\left( \sum _{m\le x}|a(m)|\right) \frac{1}{(n+x)^{\sigma }}+\sigma \int _{1}^{x}\left( \sum _{m\le t}|a(m)|\right) \frac{1}{(n+t)^{\sigma +1}}dt. \end{aligned}$$
(3)

Since \( \sum _{m=1}^{\infty }\frac{a(m)}{m^s} \) has abscissa of absolute convergence \( \alpha \), it is well known that \(\sum _{m\le x}|a(m)|=o(x^{\alpha +\epsilon })\) for any \( \epsilon >0 \). So the term \( \left( \sum _{m\le x}|a(m)|\right) \frac{1}{(n+x)^{\sigma }} \) tends to zero when x tends to \( \infty \). Therefore by taking x tends to \( \infty \) in Eq. (3), we get that

$$\begin{aligned} \sum _{m=1}^{\infty }\frac{|a(m)|}{(n+m)^{\sigma }}=\sigma \int _{1}^{\infty }\left( \sum _{m\le t}|a(m)|\right) \frac{1}{(n+t)^{\sigma +1}}dt. \end{aligned}$$
(4)

Now using the fact \( \sum _{m\le t}|a(m)|\gg t^{\alpha } \) for every \( t\ge 1\) in Eq. (4), we have

$$\begin{aligned} \sum _{m=1}^{\infty }\frac{|a(m)|}{(n+m)^{\sigma }}&\gg \sigma \int _{1}^{\infty }\frac{t^{\alpha }}{(n+t)^{\sigma +1}}dt\\&\gg \sigma \int _{n}^{\infty }\frac{t^{\alpha }}{(n+t)^{\sigma +1}}dt \gg _{\sigma }\frac{1}{n^{\sigma -\alpha }}\,. \end{aligned}$$

\(\square \)

3 Proof of Theorem 1

Proof

We will prove the theorem by induction on r.

The case \( r=1 \) easily follows from the hypothesis.

Next we will prove the case \( r=2 \). Let \( (s_1,s_2)\in U_2 \), then \( \sigma _2>\alpha _2 \) and \( \sigma _2+\sigma _1>\alpha _2+\alpha _1 \). Then we can choose \( \epsilon _2>0 \) such that

$$\begin{aligned} \sigma _2>\alpha _2+\epsilon _2\,\,\text {and}\,\, \sigma _2+\sigma _1>\alpha _2+\alpha _1+\epsilon _2\,. \end{aligned}$$
(5)

Now using the hypothesis and (5), we get that

$$\begin{aligned}&\sum _{m_1=1}^{\infty }\frac{|a_1(m_1)|}{m_1^{\sigma _1}} \sum _{m_2=1}^{\infty }\frac{|a_2(m_2)|}{(m_1+m_2)^{\sigma _2}} \\&=\sum _{m_1=1}^{\infty }\frac{| a_1(m_1)|}{m_1^{\sigma _1+\sigma _2-(\alpha _2+\epsilon _2)}} \sum _{m_2=1}^{\infty }\frac{| a_2(m_2)|}{(m_1+m_2)^{\alpha _2+\epsilon _2}}\left( \frac{m_1}{m_1+m_2}\right) ^{\sigma _2-(\alpha _2+\epsilon _2)}\\&<\left( \sum _{m_1=1}^{\infty }\frac{| a_1(m_1)|}{m_1^{\sigma _1+\sigma _2-(\alpha _2+\epsilon _2)}}\right) \left( \sum _{m_2=1}^{\infty }\frac{|a_2(m_2)|}{m_2^{\alpha _2+\epsilon _2}}\right) <\infty \,. \end{aligned}$$

Suppose the theorem is true for \( r-1,\,r\ge 3 \). Now we will prove for r. Let \( (s_1,s_2,\ldots ,s_r)\in U_r \), then

$$\begin{aligned}\sigma _r+\sigma _{r-1}+\cdots +\sigma _{r-i}> \alpha _r+\alpha _{r-1}+\cdots +\alpha _{r-i}\,\,\text {for}\,\,0\le i\le r-1. \end{aligned}$$

So there exists an \( \epsilon _r >0 \) such that

$$\begin{aligned} \sigma _r+\sigma _{r-1}+\cdots +\sigma _{r-i}> \alpha _r+\alpha _{r-1}+\cdots +\alpha _{r-i}+\epsilon _r\,\,\text {for}\,\,0\le i\le r-1. \end{aligned}$$
(6)

Now consider

$$\begin{aligned}&\sum _{m_1=1}^{\infty }\frac{|a_1(m_1)|}{m_1^{\sigma _1}}\cdots \sum _{m_{r-1}=1}^{\infty }\frac{| a_{r-1}(m_{r-1})|}{(m_1+\cdots +m_{r-1})^{\sigma _{r-1}}}\sum _{m_{r}=1}^{\infty }\frac{|a_{r}(m_{r})|}{(m_1+\cdots +m_r)^{\sigma _{r}}}\\&=\sum _{m_1=1}^{\infty }\frac{|a_1(m_1)|}{m_1^{\sigma _1}}\cdots \sum _{m_{r-2}=1}^{\infty }\frac{| a_{r-2}(m_{r-2})|}{(m_1+\cdots +m_{r-2})^{\sigma _{r-2}}}\\&\quad \sum _{m_{r-1}=1}^{\infty }\frac{| a_{r-1}(m_{r-1})|}{(m_1+\cdots +m_{r-1})^{\sigma _{r-1}+\sigma _r-(\alpha _r+\epsilon _r)}}\\&\quad \sum _{m_{r}=1}^{\infty }\frac{| a_{r}(m_{r})|}{(m_1+\cdots +m_r)^{\alpha _{r}+\epsilon _r}}\left( \frac{m_1+\cdots +m_{r-1}}{m_1+\cdots +m_r}\right) ^{\sigma _r-(\alpha _r+\epsilon _r)}\\&<\left( \sum _{m_1=1}^{\infty }\frac{|a_1(m_1)|}{m_1^{\sigma _1}}\cdots \sum _{m_{r-1}=1}^{\infty }\frac{| a_{r-1}(m_{r-1})|}{(m_1+\cdots +m_{r-1})^{\sigma _{r-1}+\sigma _r-(\alpha _r+\epsilon _r)}}\right) \\&\quad \left( \sum _{m_{r}=1}^{\infty }\frac{| a_{r}(m_{r})|}{m_r^{\alpha _{r}+\epsilon _r}}\right) <\infty \,, \end{aligned}$$

where the above inequality follows from induction hypothesis and (6). \(\square \)

4 Proof of Theorem 2

Proof

Let \(\sigma _r+\sigma _{r-1}+\cdots +\sigma _{r-i}>\alpha _r+\alpha _{r-1}+\cdots +\alpha _{r-i}\) for \(0\le i\le r-1\), then it follows from hypothesis and Theorem 1 that the series defined in (2) is absolutely convergent at \( (s_1,s_2,\ldots ,s_r) \).

We will prove the converse part by induction on r.

The case \( r=1 \) easily follows from Lemma 2.

Suppose the theorem is true for \( r-1 \), \( r\ge 2 \). Now we will prove for r. By hypothesis, we have

$$\begin{aligned}&\sum _{m_1=1}^{\infty }\sum _{m_2=1}^{\infty }\cdots \sum _{m_r=1}^{\infty } \frac{|a_1(m_1)||a_2(m_2)|\cdots |a_r(m_r)|}{m_1^{\sigma _1}(m_1+m_2)^{\sigma _2}\cdots (m_1+m_2+\cdots +m_r)^{\sigma _r}}<\infty \,.\end{aligned}$$
(7)

The above inequality implies that

$$\begin{aligned} \sum _{m_r=1}^{\infty }\frac{|a_r(m_r)|}{(m_1+m_2+\cdots +m_r)^{\sigma _r}}<\infty \,\,\text {for any}\,\,(m_1,m_2,\ldots ,m_{r-1})\in {\mathbb {N}}^{r-1} \end{aligned}$$

and then by applying Lemmas 1 and 2, we have \( \sigma _r>\alpha _r \). Now \( \text {for any}\,\,(m_1,m_2,\ldots ,m_{r-1})\in {\mathbb {N}}^{r-1}\), we get from Lemma 3 that

$$\begin{aligned} \sum _{m_r=1}^{\infty }\frac{|a_r(m_r)|}{(m_1+m_2+\cdots +m_r)^{\sigma _r}}\gg _{\sigma _r}\frac{1}{(m_1+\cdots +m_{r-1})^{\sigma _r-\alpha _r}}\,. \end{aligned}$$
(8)

Therefore from expressions (7) and (8), we have

$$\begin{aligned} \sum _{m_1=1}^{\infty }\sum _{m_2=1}^{\infty }\cdots \sum _{m_{r-1}=1}^{\infty } \frac{|a_1(m_1)||a_2(m_2)|\cdots |a_{r-1}(m_{r-1})|}{m_1^{\sigma _1}(m_1+m_2)^{\sigma _2}\cdots (m_1+\cdots +m_{r-2})^{\sigma _{r-2}}(m_1+\cdots +m_{r-1})^{\sigma _{r-1}+\sigma _r-\alpha _r}}<\infty \,.\end{aligned}$$

Then from induction hypothesis, it follows that

$$\begin{aligned} \sigma _r+\sigma _{r-1}+\cdots +\sigma _{r-i}>\alpha _r+\alpha _{r-1}+\cdots +\alpha _{r-i}\,\,\text {for}\,\,1\le i\le r-1. \end{aligned}$$

\(\square \)

5 Proof of Theorem 3

Proof

It is enough to show that for \(\sigma _r+\sigma _{r-1}+\cdots +\sigma _{r-i}\ge \sigma _r^{\prime }+\sigma _{r-1}^{\prime }+\cdots +\sigma _{r-i}^{\prime }\), \(0\le i\le r-1\), we have

$$\begin{aligned} \begin{aligned}&\sum _{m_1=1}^{\infty }\sum _{m_2=1}^{\infty }\cdots \sum _{m_r=1}^{\infty } \frac{|a_1(m_1)||a_2(m_2)|\cdots |a_r(m_r)|}{m_1^{\sigma _1}(m_1+m_2)^{\sigma _2}\cdots (m_1+m_2+\cdots +m_r)^{\sigma _r}}\\&\le \sum _{m_1=1}^{\infty }\sum _{m_2=1}^{\infty }\cdots \sum _{m_r=1}^{\infty } \frac{|a_1(m_1)||a_2(m_2)|\cdots |a_r(m_r)|}{m_1^{\sigma _1^{\prime }}(m_1+m_2)^{\sigma _2^{\prime }}\cdots (m_1+m_2+\cdots +m_r)^{\sigma _r^{\prime }}}. \end{aligned} \end{aligned}$$
(9)

Now consider

$$\begin{aligned}&\sum _{m_1=1}^{\infty }\frac{|a_1(m_1)|}{m_1^{\sigma _1}}\cdots \sum _{m_{r-1}=1}^{\infty }\frac{| a_{r-1}(m_{r-1})|}{(m_1+\cdots +m_{r-1})^{\sigma _{r-1}}}\sum _{m_{r}=1}^{\infty }\frac{|a_{r}(m_{r})|}{(m_1+\cdots +m_r)^{\sigma _{r}}}\\&=\sum _{m_1=1}^{\infty }\frac{|a_1(m_1)|}{m_1^{\sigma _1}}\cdots \sum _{m_{r-2}=1}^{\infty }\frac{| a_{r-2}(m_{r-2})|}{(m_1+\cdots +m_{r-2})^{\sigma _{r-2}}}\sum _{m_{r-1}=1}^{\infty }\frac{| a_{r-1}(m_{r-1})|}{(m_1+\cdots +m_{r-1})^{\sigma _r+\sigma _{r-1}-\sigma _r^{\prime }}}\\&\sum _{m_{r}=1}^{\infty }\frac{|a_{r}(m_{r})|}{(m_1+\cdots +m_r)^{\sigma _{r}^{\prime }}}\left( \frac{m_1+\cdots +m_{r-1}}{m_1+\cdots +m_r}\right) ^{\sigma _r-\sigma _r^{\prime }}. \end{aligned}$$

Here \(\left( \frac{m_1+\cdots +m_{r-1}}{m_1+\cdots +m_r}\right) ^{\sigma _r-\sigma _r^{\prime }}\le 1 \) as \( \sigma _r\ge \sigma _r^{\prime } \), therefore we have

$$\begin{aligned}&\sum _{m_1=1}^{\infty }\frac{|a_1(m_1)|}{m_1^{\sigma _1}}\cdots \sum _{m_{r-1}=1}^{\infty }\frac{| a_{r-1}(m_{r-1})|}{(m_1+\cdots +m_{r-1})^{\sigma _{r-1}}}\sum _{m_{r}=1}^{\infty }\frac{|a_{r}(m_{r})|}{(m_1+\cdots +m_r)^{\sigma _{r}}}\\&\le \sum _{m_1=1}^{\infty }\frac{|a_1(m_1)|}{m_1^{\sigma _1}}\cdots \sum _{m_{r-2}=1}^{\infty }\frac{| a_{r-2}(m_{r-2})|}{(m_1+\cdots +m_{r-2})^{\sigma _{r-2}}}\sum _{m_{r-1}=1}^{\infty }\frac{| a_{r-1}(m_{r-1})|}{(m_1+\cdots +m_{r-1})^{\sigma _r+\sigma _{r-1}-\sigma _r^{\prime }}}\\&\sum _{m_{r}=1}^{\infty }\frac{|a_{r}(m_{r})|}{(m_1+\cdots +m_r)^{\sigma _{r}^{\prime }}}\\&=\sum _{m_1=1}^{\infty }\frac{|a_1(m_1)|}{m_1^{\sigma _1}}\cdots \sum _{m_{r-3}=1}^{\infty }\frac{| a_{r-3}(m_{r-3})|}{(m_1+\cdots +m_{r-3})^{\sigma _{r-3}}}\sum _{m_{r-2}=1}^{\infty }\frac{| a_{r-2}(m_{r-2})|}{(m_1+\cdots +m_{r-2})^{\sigma _r+\sigma _{r-1}+\sigma _{r-2}-\sigma _r^{\prime }-\sigma _{r-1}^{\prime }}}\\&\sum _{m_{r-1}=1}^{\infty }\frac{| a_{r-1}(m_{r-1})|}{(m_1+\cdots +m_{r-1})^{\sigma _{r-1}^{\prime }}}\left( \frac{m_1+\cdots +m_{r-2}}{m_1+\cdots +m_{r-1}}\right) ^{\sigma _r+\sigma _{r-1}-\sigma _r^{\prime }-\sigma _{r-1}^{\prime }}\sum _{m_{r}=1}^{\infty }\frac{|a_{r}(m_{r})|}{(m_1+\cdots +m_r)^{\sigma _{r}^{\prime }}}\\&\le \sum _{m_1=1}^{\infty }\frac{|a_1(m_1)|}{m_1^{\sigma _1}}\cdots \sum _{m_{r-3}=1}^{\infty }\frac{| a_{r-3}(m_{r-3})|}{(m_1+\cdots +m_{r-3})^{\sigma _{r-3}}}\sum _{m_{r-2}=1}^{\infty }\frac{| a_{r-2}(m_{r-2})|}{(m_1+\cdots +m_{r-2})^{\sigma _r+\sigma _{r-1}+\sigma _{r-2}-\sigma _r^{\prime }-\sigma _{r-1}^{\prime }}}\\&\sum _{m_{r-1}=1}^{\infty }\frac{| a_{r-1}(m_{r-1})|}{(m_1+\cdots +m_{r-1})^{\sigma _{r-1}^{\prime }}}\sum _{m_{r}=1}^{\infty }\frac{|a_{r}(m_{r})|}{(m_1+\cdots +m_r)^{\sigma _{r}^{\prime }}}\,, \end{aligned}$$

where the last inequality follows from the fact \( \left( \frac{m_1+\cdots +m_{r-2}}{m_1+\cdots +m_{r-1}}\right) ^{\sigma _r+\sigma _{r-1}-\sigma _r^{\prime }-\sigma _{r-1}^{\prime }}\le 1 \) as \( \sigma _r+\sigma _{r-1}\ge \sigma _r^{\prime }+\sigma _{r-1}^{\prime } \).

Continuing in the same manner, we obtain (9). \(\square \)