Some conditions for sequences to be minimal completely monotonic

Xifeng Wang; Senlin Guo; Xifeng Wang; Senlin Guo

doi:10.3934/math.2023496

AIMS Mathematics

2023, Volume 8, Issue 4: 9832-9839. doi: 10.3934/math.2023496

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Research article

Some conditions for sequences to be minimal completely monotonic

Xifeng Wang ^,,
Senlin Guo

College of Science, Zhongyuan University of Technology, Zhengzhou, Henan 450007, China

Received: 07 November 2022 Revised: 30 January 2023 Accepted: 06 February 2023 Published: 22 February 2023
MSC : 44A60, 44A10

In this article, we establish some necessary conditions for sequences to be minimal completely monotonic. We also present some properties for completely monotonic sequences.

Keywords:

Citation: Xifeng Wang, Senlin Guo. Some conditions for sequences to be minimal completely monotonic[J]. AIMS Mathematics, 2023, 8(4): 9832-9839. doi: 10.3934/math.2023496

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Abstract

In this article, we establish some necessary conditions for sequences to be minimal completely monotonic. We also present some properties for completely monotonic sequences.

1. Introduction and main results

We will first introduce some definitions and some basic results on completely monotonic sequences.

Recall that ^[8] a sequence $\left\{\mu_{n}\right\}_{n=0}^{\infty}$ is called to be completely monotonic if

$(-1)^{k} \Delta^{k} \mu_{n} \geq 0, \quad n, k \in \mathbb{N}_{0}$

(1)

where

$\Delta^{0} \mu_{n} = \mu_{n}$

(2)

and

$\Delta^{k+1} \mu_{n} = \Delta^{k} \mu_{n+1}-\Delta^{k} \mu_{n}.$

(3)

Here, in (1) and throughout the paper, $\mathbb{N}_{0}: = \{0\} \cup \mathbb{N}$ and $\mathbb{N}$ is the set of all positive integers. Such a sequence is called totally monotonic in ^[21].

In ^[9] the authors showed that for a completely monotonic sequence $\left\{\mu_n\right\}_{n=0}^{\infty}$ we always have

$(-1)^{k} \Delta^{k} \mu_{n} > 0, \quad n, k \in \mathbb{N}_{0},$

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

Hausdorff ^[8] proved the following fundamental result, in terms of Stieltjes integrals, for completely monotonic sequences (Hausdorff Theorem): a sequence $\left\{\mu_n\right\}_{n = 0}^{\infty}$ is completely monotonic if and only if there exists a non-decreasing and bounded function $\alpha(t)$ on the interval $[0, 1]$ such that

$\mu_n = \int_0^1 t^n d \alpha(t), \quad n \in \mathbb{N}_0 .$

(4)

Also recall that ^[19] a sequence $\left\{\mu_n\right\}_{n = 0}^{\infty}$ is called minimal completely monotonic if it is completely monotonic and if it will not be completely monotonic when $\mu_0$ is replaced by a number which is less than $\mu_0$ .

In ^[5] the author proved that for each completely monotonic sequence $\left\{\mu_n\right\}_{n = 0}^{\infty}$ there exists one and only one number $\mu_0^*$ such that the sequence

$\left\{\mu_0^*, \mu_1, \mu_2, \mu_3, \cdots\right\}$

is minimal completely monotonic.

One of the results of ^[6] is that: suppose that the sequence $\left\{\mu_n\right\}_{n = 1}^{\infty}$ is completely monotonic and that the series

$\sum\limits_{j = 0}^{\infty}(-1)^j \Delta^j \mu_1$

converges. Let $\mu_0$ be such that

$\mu_0 \geq \sum\limits_{j = 0}^{\infty}(-1)^j \Delta^j \mu_1 .$

Then the sequence

$\left\{\mu_n\right\}_{n = 0}^{\infty}$

is completely monotonic.

In ^[15] the following result, among others, was established. Suppose that the sequence $\left\{\mu_n\right\}_{n = 1}^{\infty}$ is completely monotonic and that the series

$\sum\limits_{j = 0}^{\infty}(-1)^j \Delta^j \mu_1$

(5)

converges. Let

$\mu_0^*: = \sum\limits_{j = 0}^{\infty}(-1)^j \Delta^j \mu_1 .$

(6)

Then the sequence

$\left\{\mu_{0}^{*}, \mu_{1}, \mu_{2}, \mu_{3}, \cdots\right\}$

is minimal completely monotonic.

We also note that in ^[15] the authors gave a counterexample showing that complete monotonicity of the sequence $\left\{\mu_n\right\}_{n = 1}^{\infty}$ cannot guarantee the convergence of the series

$\sum\limits_{j = 0}^{\infty}(-1)^j \Delta^j \mu_1.$

(7)

A function f is said ^[1] 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 $n \in \mathbb{N}_0$

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

(8)

In ^[10] the authors showed that if the function $f$ is completely monotonic on the interval $[a, \infty)$ , the sequence $\left\{\Delta x_k\right\}_{k = 0}^{\infty}$ is completely monotonic and $x_0 \geq a$ , then the sequence $\left\{f\left(x_k\right)\right\}_{k = 0}^{\infty}$ is completely monotonic.

The following result was obtained in ^[5]. Suppose that the sequence $\left\{\mu_n\right\}_{n = 0}^{\infty}$ is completely monotonic. Then for any $\varepsilon \in(0, 1)$ , there exists a continuous interpolating function $f(x)$ on the interval $[0, \infty)$ such that $\left.f\right|_{[0, \varepsilon]}$ and $\left.f\right|_{[ \varepsilon, \infty)}$ are both completely monotonic and $f(n) = \mu_n, n \in \mathbb{N}_0$

For the operation of pointwise convergence, the author ^[20] showed that suppose that for $n \in \mathbb{N}$ , the function $f_n$ is completely monotonic on the interval I, where $I = (a, b)$ or $(a, b]$ . If the limit function

$f(x) = \lim\limits _{n \rightarrow \infty} f_n(x)$

exists on the interval I, then $f$ is completely monotonic on I.

Here we would like to point out that in the result above the interval I cannot be $[a, b)$ or $[a, b]$ . For example, let

$f_n(x) = \frac{1}{x^n}, \quad n \in \mathbb{N}$

and $I_1 = [1, b)$ or $I_1 = [1, b]$ .

It is easy to verify that

$f_n^{(k)}(x)=\frac{(n+k-1) !}{(n-1) !} \frac{(-1)^k}{x^{n+k}}, \quad k \in \mathbb{N}_0.$

Hence $f_n(x)$ is completely monotonic on the interval $I_1$ . The limit function

$f(x) = \lim\limits _{n \rightarrow \infty} f_n(x) = \begin{cases}1, & x = 1 \\ 0, & x \in I_1-\{1\} .\end{cases}$

Clearly the function $f$ is not completely monotonic on the interval $I_1$ because $f$ is not continuous on $I_1$ .

There is rich literature on completely monotonic functions and completely monotonic sequences, and their applications. For more recent works, see, for example, ^{[2,3,4,5,6,7,11,12,13,14,15,16,17,18]}.

In this article, we shall further investigate completely monotonic and minimal completely monotonic sequences. By using the Hausdorff Theorem, some necessary conditions for sequences to be completely monotonic are presented and proved. Also some properties of minimal completely monotonic sequences are established. The main results of the article are as follows.

Theorem 1. Suppose that the sequence $\left\{\mu_n\right\}_{n = 0}^{\infty}$ is completely monotonic. Then, for any $k \in \mathbb{N}_0$ and any $m \in \mathbb{N}$ , the series

$\sum\limits_{j = k}^{\infty}(-1)^j \Delta^j \mu_{m+1}$

converges and

$(-1)^k \Delta^k \mu_m = \sum\limits_{j = k}^{\infty}(-1)^j \Delta^j \mu_{m+1} .$

(9)

Theorem 2. Suppose that the sequence $\left\{\mu_n\right\}_{n = 0}^{\infty}$ is completely monotonic. Then, for any $m \in \mathbb{N}$ , the series

$\sum\limits_{j = 0}^{\infty}(-1)^j \Delta^j \mu_{m+1}$

converges and

$\mu_m = \sum\limits_{j = 0}^{\infty}(-1)^j \Delta^j \mu_{m+1} .$

(10)

Theorem 3. Suppose that the sequence $\left\{\mu_n\right\}_{n = 0}^{\infty}$ is minimal completely monotonic. Then, for any $m \in \mathbb{N}_0$ , the series

$\sum\limits_{j = 0}^{\infty}(-1)^j \Delta^j \mu_{m+1}$

converges and

$\mu_m = \sum\limits_{j = 0}^{\infty}(-1)^j \Delta^j \mu_{m+1} .$

(11)

Theorem 4. Suppose that the sequence $\left\{\mu_n\right\}_{n = 0}^{\infty}$ is minimal completely monotonic. Then, for any $k, m \in \mathbb{N}_0$ , the series

$\sum\limits_{j = k}^{\infty}(-1)^j \Delta^j \mu_{m+1}$

converges and

$(-1)^k \Delta^k \mu_m = \sum\limits_{j = k}^{\infty}(-1)^j \Delta^j \mu_{m+1} .$

(12)

2. Lemmas

Lemma 1. If the sequence $\left\{\mu_n\right\}_{n = 0}^{\infty}$ is completely monotonic, then for all $i, n \in \mathbb{N}_0$

$(-1)^i \Delta^i \mu_n = \int_0^1(1-t)^i t^n d \alpha(t),$

(13)

where $\alpha(t)$ is a non-decreasing and bounded function on the interval $[0, 1]$ .

Proof of Lemma 1. Let $n \in \mathbb{N}_0$ . When $i = 0$ by (2) and the Haudorff Theorem, we see that (13) is true. Now suppose that (13) is true for $i \in \mathbb{N}_0$ By (3) and (4) we have

$\begin{aligned} (-1)^{i+1} \Delta^{i+1} \mu_n & = (-1)^{i+1}\left(\Delta^i \mu_{n+1}-\Delta^i \mu_n\right) \\ & = \int_0^1(1-t)^i t^n d \alpha(t)-\int_0^1(1-t)^i t^{n+1} d \alpha(t) \\ & = \int_0^1(1-t)^{i+1} t^n d \alpha(t), \end{aligned}$

which means that (13) is also true for $i+1$ By induction we see that (13) is true for all $i, n \in \mathbb{N}_0$ The proof of Lemma 1 is completed.

Lemma 2 (^[15]). Suppose that the sequence $\left\{\mu_n\right\}_{n = 0}^{\infty}$ is minimal completely monotonic. Then the series

$\sum\limits_{j = 0}^{\infty}(-1)^j \Delta^j \mu_1$

is convergent and

$\mu_0 = \sum\limits_{j = 0}^{\infty}(-1)^{j^j \Delta^j \mu_1 .} .$

3. Proof of the results

We now prove the main results of the paper.

Proof of Theorem 1. For $k \in \mathbb{N}_0$ and $m \in \mathbb{N}$ , by Lemma 1, we have

$\begin{aligned} & (-1)^k \Delta^k \mu_m-\sum\limits_{j = k}^n(-1)^j \Delta^j \mu_{m+1} \\ = & \int_0^1(1-t)^k t^m d \alpha(t)-\sum\limits_{j = k}^n \int_0^1(1-t)^j t^{m+1} d \alpha(t) \\ = & \int_0^1(1-t)^k t^m d \alpha(t)-\int_0^1 \sum\limits_{j = k}^n(1-t)^j \cdot t^{m+1} d \alpha(t) \\ = & \int_0^1(1-t)^k t^m d \alpha(t)-\int_0^1 \frac{(1-t)^k-(1-t)^{n+1}}{t} \cdot t^{m+1} d \alpha(t) \\ = & \int_0^1 t(1-t)^{n+1} t^{m-1} d \alpha(t), \quad n > k . \end{aligned}$

(14)

Since the function $t(1-t)^{n+1}$ attains its maximum on the interval $[0, 1]$ at $1 /(n+2)$ , we have

$0 \leq \int_0^1 t(1-t)^{n+1} t^{m-1} d \alpha(t) \leq \frac{1}{n+2}\left(1-\frac{1}{n+2}\right)^{n+1} \int_0^1 t^{m-1} d \alpha(t) .$

By the Hausdorff Theorem we obtain

$0 \leq \int_0^1 t(1-t)^{n+1} t^{m-1} d \alpha(t) \leq \frac{1}{n+2}\left(1-\frac{1}{n+2}\right)^{n+1} \mu_{m-1} .$

Since

$\left(1-\frac{1}{n+2}\right)^{n+1} \rightarrow 1 / e \quad \text { as } n \rightarrow \infty,$

we have

$\lim\limits _{n \rightarrow \infty} \frac{1}{n+2}\left(1-\frac{1}{n+2}\right)^{n+1} \mu_{m-1} = 0 .$

Consequently,

$\lim\limits _{n \rightarrow \infty} \int_0^1 t(1-t)^{n+1} t^{m-1} d \alpha(t) = 0 .$

Then from (14) we obtain

$\lim\limits _{n \rightarrow \infty}\left((-1)^k \Delta^k \mu_m-\sum\limits_{j = k}^n(-1)^j \Delta^j \mu_{m+1}\right) = 0 .$

(15)

That is,

$\lim\limits _{n \rightarrow \infty} \sum\limits_{j = k}^n(-1)^j \Delta^j \mu_{m+1} = (-1)^k \Delta^k \mu_m .$

(16)

Therefore the series

$\sum\limits_{j = k}^{\infty}(-1)^j \Delta^j \mu_{m+1}$

converges and

$(-1)^k \Delta^k \mu_m = \sum\limits_{j = k}^{\infty}(-1)^j \Delta^j \mu_{m+1} .$

(17)

The proof of Theorem 1 is completed.

Proof of Theorem 2. Let $k$ be zero. Then from Theorem 1 we can acquire the conclusion.

The proof of Theorem 2 is hence completed.

Proof of Theorem 3. If $m = 0$ then from Lemma 2 we can obtain the conclusion. If $m \in \mathbb{N}$ in view that a minimal completely monotonic sequence is also completely monotonic, then from Theorem 2 we can obtain the conclusion.

The proof of Theorem 3 is thus completed.

Proof of Theorem 4. Let $m$ be a fixed non-negative integer. Then from Theorem 3 we see that

$\mu_m = \sum\limits_{j = 0}^{\infty}(-1)^j \Delta^j \mu_{m+1},$

(18)

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

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

$\begin{aligned} (-1)^{r+1} \Delta^{r+1} \mu_m & = (-1)^{r+1}\left(\Delta^r \mu_{m+1}-\Delta^r \mu_m\right) \\ & = \sum\limits_{j = r}^{\infty}(-1)^j \Delta^j \mu_{m+1}-(-1)^r \Delta^r \mu_{m+1} \\ & = \sum\limits_{j = r+1}^{\infty}(-1)^j \Delta^j \mu_{m+1}, \end{aligned}$

(19)

which means that (12) is also valid for $k = r+1$ . Therefore by induction, (12) is valid for all $k \in \mathbb{N}_0$ .

The proof of Theorem 4 is thus completed.

4. Conclusions

In this article, we presented some necessary conditions for a sequence to be a completely monotonic sequence. We also established some necessary conditions for a sequence to be minimal completely monotonic. The Hausdorff Theorem plays the key role when we prove our results.

Acknowledgments

The authors thank the editor and the referees for their valuable suggestions which significantly improve the quality of this paper.

Conflict of interest

The authors declare that they have no competing interests.

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AIMS Mathematics

Some conditions for sequences to be minimal completely monotonic

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Abstract

1. Introduction and main results

2. Lemmas

3. Proof of the results

4. Conclusions

Acknowledgments

Conflict of interest

References

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Abstract

1. Introduction and main results

2. Lemmas

3. Proof of the results

4. Conclusions

Acknowledgments

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AIMS Mathematics

Some conditions for sequences to be minimal completely monotonic

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Abstract

1. Introduction and main results

2. Lemmas

3. Proof of the results

4. Conclusions

Acknowledgments

Conflict of interest

References

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1. Introduction and main results

2. Lemmas

3. Proof of the results

4. Conclusions

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