A certain class of completely monotonic sequences

Abstract

In this article, we present some necessary conditions, a sufficient condition and a necessary and sufficient condition for sequences to be completely monotonic. One counterexample is also presented.

MSC:40A05, 26A45, 26A48, 39A60.

1 Introduction and the main results

We first recall some definitions and basic results on or related to completely monotonic sequences and completely monotonic functions.

Definition 1 [1]

A sequence ${\{{\mu }_n\}}_{n=0}^{\mathrm{\infty }}$ is called a moment sequence if there exists a function $\alpha (t)$ of bounded variation on the interval $[0,1]$ such that

$${\mu }_n={\int }_0^1{t}^{n}d\alpha (t),n\in {\mathbb{N}}_0.$$

(1)

Here, in Definition 1 and throughout the paper,

$${\mathbb{N}}_0:=\{0\}\cup \mathbb{N},$$

and ℕ is the set of all positive integers.

Definition 2 [1]

A sequence ${\{{\mu }_n\}}_{n=0}^{\mathrm{\infty }}$ is called completely monotonic if

$${(-1)}^k{\mathrm{\Delta }}^k{\mu }_n\geqq 0,n,k\in {\mathbb{N}}_0,$$

(2)

where

$${\mathrm{\Delta }}^0{\mu }_n={\mu }_n$$

(3)

and

$${\mathrm{\Delta }}^{k+1}{\mu }_n={\mathrm{\Delta }}^k{\mu }_{n+1}-{\mathrm{\Delta }}^k{\mu }_n.$$

(4)

Such a sequence is called totally monotone in [2].

From Definition 2, using mathematical induction, we can prove, for a completely monotonic sequence ${\{{\mu }_n\}}_{n=0}^{\mathrm{\infty }}$, that the sequence ${\{{(-1)}^m{\mathrm{\Delta }}^m{\mu }_n\}}_{n=0}^{\mathrm{\infty }}$ is non-increasing for any fixed $m\in {\mathbb{N}}_0$, and that the sequence ${\{{(-1)}^m{\mathrm{\Delta }}^m{\mu }_n\}}_{m=0}^{\mathrm{\infty }}$ is non-increasing for any fixed $n\in {\mathbb{N}}_0$. The difference equation (4) plays an important role in the proofs of these properties and our main results of this paper.

In [3], the authors showed that for a completely monotonic sequence ${\{{\mu }_n\}}_{n=0}^{\mathrm{\infty }}$, we always have

$${(-1)}^k{\mathrm{\Delta }}^k{\mu }_n>0,n,k\in {\mathbb{N}}_0,$$

(5)

unless ${\mu }_n=c$, a constant for all $n\in \mathbb{N}$.

Let

$${\lambda }_{k,m}:=\left(\genfrac{}{}{0ex}{}{k}{m}\right){(-1)}^{k-m}{\mathrm{\Delta }}^{k-m}{\mu }_m,k,m\in {\mathbb{N}}_0.$$

(6)

It was shown (see [1]) as follows.

Theorem 1 A sequence ${\{{\mu }_n\}}_{n=0}^{\mathrm{\infty }}$ is a moment sequence if and only if there exists a constant L such that

$$\sum _{m=0}^k|{\lambda }_{k,m}|<L,k\in {\mathbb{N}}_0,$$

(7)

where in (7), ${\lambda }_{k,m}$ is defined by (6).

For completely monotonic sequences, the following is the well-known Hausdorff’s theorem (see [1]).

Theorem 2 A sequence ${\{{\mu }_n\}}_{n=0}^{\mathrm{\infty }}$ is completely monotonic if and only if there exists a non-decreasing and bounded function $\alpha (t)$ on $[0,1]$ such that

$${\mu }_n={\int }_0^1{t}^{n}d\alpha (t),n\in {\mathbb{N}}_0.$$

(8)

From this theorem, we know (see [1]) that a completely monotonic sequence is a moment sequence and is as follows.

Theorem 3 A necessary and sufficient condition that the sequence ${\{{\mu }_n\}}_{n=0}^{\mathrm{\infty }}$ should be a moment sequence is that it should be the difference of two completely monotonic sequences.

We also recall the following definition.

Definition 3 [1]

A function f is said to be completely monotonic on an interval I if f is continuous on I has derivatives of all orders on $I^o$ (the interior of I) and for all $n\in {\mathbb{N}}_0$,

$${(-1)}^n{f}^{(n)}(x)\geqq 0,x\in I^o.$$

(9)

Some mathematicians use the terminology completely monotone instead of completely monotonic. The class of all completely monotonic functions on the interval I is denoted by $CM(I)$.

The completely monotonic functions and completely monotonic sequences have remarkable applications in probability and statistics [4–10], physics [11, 12], numerical and asymptotic analysis [2], etc.

For the completely monotonic functions on the interval $[0,\mathrm{\infty })$, Widder proved (see [1]).

Theorem 4 A function f on the interval $[0,\mathrm{\infty })$ is completely monotonic if and only if there exists a bounded and non-decreasing function $\alpha (t)$ on $[0,\mathrm{\infty })$ such that

$$f(x)={\int }_0^{\mathrm{\infty }}{e}^{-xt}d\alpha (t).$$

(10)

There is rich literature on completely monotonic functions. For more recent works, see, for example, [13–26].

There exists a close relationship between completely monotonic functions and completely monotonic sequences. For example, Widder [27] showed the following.

Theorem 5 Suppose that $f\in CM[a,\mathrm{\infty })$, then for any $\delta \geqq 0$, the sequence ${\{f(a+n\delta )\}}_{n=0}^{\mathrm{\infty }}$ is completely monotonic.

This result was generalized in [28] as follows.

Theorem 6 Suppose that $f\in CM[a,\mathrm{\infty })$. If the sequence ${\{\mathrm{\Delta }{x}_k\}}_{k=0}^{\mathrm{\infty }}$ is completely monotonic and $x_0\geqq a$, then the sequence ${\{f(x_k)\}}_{k=0}^{\mathrm{\infty }}$ is also completely monotonic.

For the meaning of $\mathrm{\Delta }{x}_k$, $k\in {\mathbb{N}}_0$ in Theorem 6, see (3) and (4).

Suppose that $f\in CM[0,\mathrm{\infty })$. By Theorem 5, we know that ${\{f(n)\}}_{n=0}^{\mathrm{\infty }}$ is completely monotonic.

The following result was obtained in [16].

Theorem 7 Suppose that the sequence ${\{{\mu }_n\}}_{n=0}^{\mathrm{\infty }}$ is completely monotonic, then for any $\epsilon \in (0,1)$, there exists a continuous interpolating function $f(x)$ on the interval $[0,\mathrm{\infty })$ such that $f{|}_{[0,\epsilon ]}$ and $f{|}_{[\epsilon ,\mathrm{\infty })}$ are both completely monotonic and

$$f(n)={\mu }_n,n\in {\mathbb{N}}_0.$$

From this result or Theorem 2, we can get the following.

Theorem 8 Suppose that the sequence ${\{{\mu }_n\}}_{n=0}^{\mathrm{\infty }}$ is completely monotonic. Then there exists a completely monotonic interpolating function $g(x)$ on the interval $[1,\mathrm{\infty })$ such that

$$g(n)={\mu }_n,n\in \mathbb{N}.$$

It should be noted that (see [[1], Chapter IV]) under the condition of Theorem 8, we cannot guarantee that there exists a completely monotonic interpolating function $g(x)$ on the interval $[0,\mathrm{\infty })$ such that

$$g(n)={\mu }_n,n\in {\mathbb{N}}_0.$$

In this article, we shall further investigate the properties of the completely monotonic sequences. We shall give some necessary conditions, a sufficient condition and a necessary and sufficient condition for sequences to be completely monotonic. More precisely we have the following results.

Theorem 9 Suppose that the sequence ${\{{\mu }_n\}}_{n=0}^{\mathrm{\infty }}$ is completely monotonic. Then, for any $m\in {\mathbb{N}}_0$, the series

$$\sum _{j=0}^{\mathrm{\infty }}{(-1)}^j{\mathrm{\Delta }}^j{\mu }_{m+1}$$

converges and

$${\mu }_m\geqq \sum _{j=0}^{\mathrm{\infty }}{(-1)}^j{\mathrm{\Delta }}^j{\mu }_{m+1}.$$

Corollary 1 Suppose that the sequence ${\{{\mu }_n\}}_{n=0}^{\mathrm{\infty }}$ is completely monotonic. Then for $m,k\in {\mathbb{N}}_0$,

$${\mu }_m={(-1)}^{k+1}{\mathrm{\Delta }}^{k+1}{\mu }_m+\sum _{i=0}^k{(-1)}^i{\mathrm{\Delta }}^i{\mu }_{m+1}.$$

(11)

Remark 1 Although from the complete monotonicity of the sequence ${\{{\mu }_n\}}_{n=0}^{\mathrm{\infty }}$, we can deduce that for any $m\in {\mathbb{N}}_0$, the series

$$\sum _{j=0}^{\mathrm{\infty }}{(-1)}^j{\mathrm{\Delta }}^j{\mu }_{m+1}$$

converges, it cannot guarantee the convergence of the series

$$\sum _{j=0}^{\mathrm{\infty }}{(-1)}^j{\mathrm{\Delta }}^j{\mu }_0.$$

For example, let

$${\mu }_n=\frac{1}{n+1},n\in {\mathbb{N}}_0.$$

Since the function

$$f(x)=\frac{1}{x+1}$$

is completely monotonic on the interval $[0,\mathrm{\infty })$, by Theorem 5, we see that the sequence

$${\{{\mu }_n\}}_{n=0}^{\mathrm{\infty }}:={\{f(n)\}}_{n=0}^{\mathrm{\infty }}={\left\{\frac{1}{n+1}\right\}}_{n=0}^{\mathrm{\infty }}$$

is completely monotonic. This conclusion can also be obtained by setting

$$\alpha (t)=t$$

in Theorem 2.

We can verify that

$${\mathrm{\Delta }}^j{\mu }_0=\frac{{(-1)}^j}{j+1}.$$

Hence,

$$\sum _{j=0}^{\mathrm{\infty }}{(-1)}^j{\mathrm{\Delta }}^j{\mu }_0=\sum _{j=0}^{\mathrm{\infty }}\frac{1}{j+1}$$

is the famous harmonic series, which is divergent.

Theorem 10 Suppose that the sequence ${\{{\mu }_n\}}_{n=0}^{\mathrm{\infty }}$ is completely monotonic. Then for any $k,m\in {\mathbb{N}}_0$,

$${(-1)}^k{\mathrm{\Delta }}^k{\mu }_m\geqq \sum _{j=k}^{\mathrm{\infty }}{(-1)}^j{\mathrm{\Delta }}^j{\mu }_{m+1}.$$

(12)

Theorem 11 Suppose that the sequence ${\{{\mu }_n\}}_{n=1}^{\mathrm{\infty }}$ is completely monotonic and that the series

$$\sum _{j=0}^{\mathrm{\infty }}{(-1)}^j{\mathrm{\Delta }}^j{\mu }_1$$

converges. Let ${\mu }_0$ be such that

$${\mu }_0\geqq \sum _{j=0}^{\mathrm{\infty }}{(-1)}^j{\mathrm{\Delta }}^j{\mu }_1.$$

Then the sequence ${\{{\mu }_n\}}_{n=0}^{\mathrm{\infty }}$ is completely monotonic.

Theorem 12 A necessary and sufficient condition for the sequence ${\{{\mu }_n\}}_{n=0}^{\mathrm{\infty }}$ to be completely monotonic is that the sequence ${\{{\mu }_n\}}_{n=1}^{\mathrm{\infty }}$ is completely monotonic, the series

$$\sum _{j=0}^{\mathrm{\infty }}{(-1)}^j{\mathrm{\Delta }}^j{\mu }_1$$

converges and

$${\mu }_0\geqq \sum _{j=0}^{\mathrm{\infty }}{(-1)}^j{\mathrm{\Delta }}^j{\mu }_1.$$

2 Proofs of the main results

Now, we are in a position to prove the main results.

Proof of Theorem 9 Since ${\{{\mu }_n\}}_{n=0}^{\mathrm{\infty }}$ is completely monotonic, by Theorem 2, there exists a non-decreasing and bounded function $\alpha (t)$ on the interval $[0,1]$ such that

$${\mu }_n={\int }_0^1{t}^{n}d\alpha (t),n\in {\mathbb{N}}_0.$$

(13)

From (3), (4) and (13), we can prove that

$${(-1)}^i{\mathrm{\Delta }}^i{\mu }_n={\int }_0^1{(1-t)}^i{t}^{n}d\alpha (t),i,n\in {\mathbb{N}}_0.$$

(14)

Now, for $k\in \mathbb{N}$, we have

$$\begin{array}{rl}\sum _{i=0}^{k-1}{(-1)}^i{\mathrm{\Delta }}^i{\mu }_{m+1}& =\sum _{i=0}^{k-1}{\int }_0^1{(1-t)}^i{t}^{m+1}d\alpha (t)\\ ={\int }_0^1{t}^{m+1}\sum _{i=0}^{k-1}{(1-t)}^{i}d\alpha (t)\\ ={\int }_0^1{t}^m(1-{(1-t)}^k)d\alpha (t)\\ ={\int }_0^1{t}^{m}d\alpha (t)-{\int }_0^1{(1-t)}^k{t}^{m}d\alpha (t)\\ ={\mu }_m-{(-1)}^k{\mathrm{\Delta }}^k{\mu }_m,m\in {\mathbb{N}}_0.\end{array}$$

Hence, for $k\in \mathbb{N}$,

$${\mu }_m={(-1)}^k{\mathrm{\Delta }}^k{\mu }_m+\sum _{i=0}^{k-1}{(-1)}^i{\mathrm{\Delta }}^i{\mu }_{m+1},m\in {\mathbb{N}}_0.$$

(15)

Since

$${(-1)}^i{\mathrm{\Delta }}^i{\mu }_n\geqq 0,i,n\in {\mathbb{N}}_0,$$

(16)

from (15), we get, for $k\geqq 1$,

$${\mu }_m\geqq \sum _{i=0}^{k-1}{(-1)}^i{\mathrm{\Delta }}^i{\mu }_{m+1},m\in {\mathbb{N}}_0.$$

(17)

From (16), we also know that

$$\sum _{j=0}^{\mathrm{\infty }}{(-1)}^j{\mathrm{\Delta }}^j{\mu }_{m+1},m\in {\mathbb{N}}_0$$

is a positive series. Then by (17), we obtain that

$$\sum _{j=0}^{\mathrm{\infty }}{(-1)}^j{\mathrm{\Delta }}^j{\mu }_{m+1},m\in {\mathbb{N}}_0$$

converges and that

$${\mu }_m\geqq \sum _{j=0}^{\mathrm{\infty }}{(-1)}^j{\mathrm{\Delta }}^j{\mu }_{m+1},m\in {\mathbb{N}}_0.$$

(18)

The proof of Theorem 9 is thus completed. □

Proof of Corollary 1 This corollary can be obtained from (15). □

Proof of Theorem 10 Let m be a fixed non-negative integer.

From Theorem 9, we see that

$${\mu }_m\geqq \sum _{j=0}^{\mathrm{\infty }}{(-1)}^j{\mathrm{\Delta }}^j{\mu }_{m+1},$$

(19)

which means that (12) is valid for $k=0$.

Suppose that (12) is valid for $k=r$. Then

$$\begin{array}{rl}{(-1)}^{r+1}{\mathrm{\Delta }}^{r+1}{\mu }_m& ={(-1)}^{r+1}({\mathrm{\Delta }}^r{\mu }_{m+1}-{\mathrm{\Delta }}^r{\mu }_m)\\ ={(-1)}^r({\mathrm{\Delta }}^r{\mu }_m-{\mathrm{\Delta }}^r{\mu }_{m+1})\\ ={(-1)}^r{\mathrm{\Delta }}^r{\mu }_m-{(-1)}^r{\mathrm{\Delta }}^r{\mu }_{m+1}\\ \geqq \sum _{j=r}^{\mathrm{\infty }}{(-1)}^j{\mathrm{\Delta }}^j{\mu }_{m+1}-{(-1)}^r{\mathrm{\Delta }}^r{\mu }_{m+1}\\ =\sum _{j=r+1}^{\mathrm{\infty }}{(-1)}^j{\mathrm{\Delta }}^j{\mu }_{m+1},\end{array}$$

(20)

which means that (12) is valid for $k=r+1$. Therefore, by mathematical induction, (12) is valid for all $k\in {\mathbb{N}}_0$. The proof of Theorem 10 is completed. □

Proof of Theorem 11 By the definition of completely monotonic sequence, we only need to prove that

$${(-1)}^k{\mathrm{\Delta }}^k{\mu }_0\geqq 0,k\in {\mathbb{N}}_0.$$

(21)

We first prove that

$${(-1)}^k{\mathrm{\Delta }}^k{\mu }_0\geqq \sum _{j=k}^{\mathrm{\infty }}{(-1)}^j{\mathrm{\Delta }}^j{\mu }_1,k\in {\mathbb{N}}_0.$$

(22)

From the condition of Theorem 11, (22) is valid for $k=0$.

Suppose that (22) is valid for $k=m$. Then we have

$$\begin{array}{rl}{(-1)}^{m+1}{\mathrm{\Delta }}^{m+1}{\mu }_0& ={(-1)}^{m+1}({\mathrm{\Delta }}^m{\mu }_1-{\mathrm{\Delta }}^m{\mu }_0)\\ ={(-1)}^m({\mathrm{\Delta }}^m{\mu }_0-{\mathrm{\Delta }}^m{\mu }_1)\\ ={(-1)}^m{\mathrm{\Delta }}^m{\mu }_0-{(-1)}^m{\mathrm{\Delta }}^m{\mu }_1\\ \geqq \sum _{j=m}^{\mathrm{\infty }}{(-1)}^j{\mathrm{\Delta }}^j{\mu }_1-{(-1)}^m{\mathrm{\Delta }}^m{\mu }_1\\ =\sum _{j=m+1}^{\mathrm{\infty }}{(-1)}^j{\mathrm{\Delta }}^j{\mu }_1,\end{array}$$

(23)

which means that (22) is valid for $k=m+1$. Therefore, by mathematical induction, (22) is valid for all $k\in {\mathbb{N}}_0$.

Since

$$\sum _{j=0}^{\mathrm{\infty }}{(-1)}^j{\mathrm{\Delta }}^j{\mu }_1$$

is a convergent positive series, we know that

$$\sum _{j=k}^{\mathrm{\infty }}{(-1)}^j{\mathrm{\Delta }}^j{\mu }_1\geqq 0,k\in {\mathbb{N}}_0.$$

(24)

From (22) and (24), we obtain that

$${(-1)}^k{\mathrm{\Delta }}^k{\mu }_0\geqq 0,k\in {\mathbb{N}}_0.$$

The proof of Theorem 11 is completed. □

Proof of Theorem 12 By Definition 2 and by setting $m=0$ in Theorem 9, we see that the condition is necessary. By Theorem 11, we know that the condition is sufficient. The proof of Theorem 12 is completed. □