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BY 4.0 license Open Access Published by De Gruyter Open Access August 4, 2020

Asymptotic relations for the products of elements of some positive sequences

  • Agata Chmielowska , Michał Różański , Barbara Smoleń EMAIL logo , Ireneusz Sobstyl and Roman Wituła
From the journal Open Mathematics

Abstract

The aim of this study was to present a simple method for finding the asymptotic relations for products of elements of some positive real sequences. The main reason to carry out this study was the result obtained by Alzer and Sandor concerning an estimation of a sequence of the product of the first k primes.

MSC 2010: 41A60; 40A20; 40A25; 11A41

1 Introduction

Let p i be the ith prime number. Let us denote p k i = 1 k p i . Consider the sequence b k k 2 k p k . Alzer and Sándor [1] proved the following inequality:

exp k ( c 0 log log k ) b k exp k ( c 1 log log k )

for every k 5 , where constants c 0 and c 1 are equal to

c 0 = 1 5 log 23 + log log 5 1.10298 ,

c 1 = 1 192 log 36 864 192 + log log 192 1 192 log ( p 192 ) 2.04287 ,

respectively. In this study, another estimation of the sequence { b k } connected with the limit lim k f ( k ) = 2 will be shown, where

f ( k ) = 1 k log k 2 k + log log k 1 k θ ( p k ) , θ ( x ) p x log p ,

and p runs over all prime numbers less than or equal to x. The above limit was proven by Alzer and Sándor [1]. Nevertheless, we decided to obtain a new proof of this limit as the original proof was obtained in an over complicated way (in a certain sense). In consequence, we also found a new result (Theorem 1), which gives a simple and universal tool for generating asymptotic relations of many known sequences of real numbers, especially for the products of elements of certain sequences.

2 Main result

The discussion is based on the following well-known fact, which is connected to d′ Alembert’s ratio test.

Lemma 1

Suppose that { a k } k = 1 is a sequence of positive terms and there exists a finite limit lim k a k + 1 a k = g . If g < 1 , then lim k a k = 0 , and if g > 1 , then lim k a k = .

This result will be used for finding the estimation of sequence { b k } k = 1 in the following way. If we can find sequences { m k } k = 1 , { M k } k = 1 of positive real numbers such that

lim k b k + 1 m k + 1 b k m k 1 > 1 , lim k b k + 1 M k + 1 b k M k 1 < 1 ,

then lim k b k m k = , lim k b k M k = 0 , thereby m k b k M k for sufficiently large k.

Using Stirling’s approximation we can write

k 2 k = ( k 2 ) ! k ! ( k 2 k ) ! 1 2 π ( k 1 ) k 1 2 1 1 k k 2 1 2 π ( k e ) k 1 2 .

From the prime number theorem, we get p k k log k . Therefore, we have

b k + 1 b k = ( k + 1 ) 2 k + 1 p k k 2 k p k 1 1 p k + 1 ( k + 1 ) k + 1 2 e k + 1 2 k k 1 2 e k 1 2 e 2 log ( k + 1 ) .

If we denote c k ( t ) = e k t ( log k ) k , then we get c k ( t ) c k + 1 ( t ) e t log ( k + 1 ) , and thereby

b k + 1 c k + 1 ( t ) b k c k ( t ) 1 e 2 t ,

where e 2 t > 1 if t < 2 and e 2 t < 1 if t > 2 . Then, for every t 1 < 2 < t 2 and for sufficiently large k the following inequalities hold

e k t 1 ( log k ) k = c k ( t 1 ) b k c k ( t 2 ) = e k t 2 ( log k ) k ,

hence

exp [ k ( t 1 log log k ) ] k 2 k p k exp [ k ( t 2 log log k ) ] .

Corollary 1

We have (see [1])

lim k log k 2 k k log p k k + log log k = 2 ,

or equivalently

lim k k 2 k p k k log k = e 2 .

In a similar way, we may find the estimation of sequence p k k = 1 . We have

p k + 1 p k = p k + 1 ( k + 1 ) log ( k + 1 ) .

Let us choose the sequence d k ( e t k log k ) k . Then, d k + 1 ( t ) d k ( t ) e t + 1 ( k + 1 ) log ( k + 1 ) and

p k + 1 d k + 1 ( t ) p k d k ( t ) 1 e t 1 ,

where e t 1 > 1 if t < 1 and e t 1 < 1 if t > 1 . Therefore, we obtain

exp k ( t 3 + log k + log log k ) p k exp k ( t 4 + log k + log log k )

for any t 3 < 1 < t 4 and for sufficiently large k. Moreover, we also have

lim k log p k k log k log log k = 1 ,

i.e.,

lim k p k k k log k = 1 e .

The last result reminds another known limit

(1) lim k k ! k k = 1 e ,

which is not an incident and comes from the general relationship presented in Theorem 1.

From now on we will use the symbol x k to denote product i = 1 k x i , where { x i } i = 1 is any real sequence. Now, we will prove our main result.

Theorem 1

Let { f k } k = 1 be an increasing sequence of positive reals. Suppose that there exists r , w and a polynomial p [ k ] such that

f k e r k p ( k ) log w k .

Then, for every pair t 1 , t 2 of real numbers satisfying the condition t 1 < r p ( 0 ) < t 2 , the following inequality holds

( e t 1 k p ( 0 ) log w k ) k f k k p ( k ) p ( 0 ) ( e t 2 k p ( 0 ) log w k ) k ,

for all sufficiently large k, which implies the relation

lim k log f k k p ( k ) p ( 0 ) k p ( 0 ) log k log ( log w k ) = r p ( 0 ) ,

or the equivalent one

(2) lim k f k k p ( k ) p ( 0 ) k k p ( 0 ) log w k = e r p ( 0 ) .

Proof

Let us set g k e t k k k p ( 0 ) log k w k . Then, we obtain

g k g k + 1 = e t 1 + 1 k k p ( 0 ) ( k + 1 ) p ( 0 ) 1 + log 1 + 1 k log k k w log w ( k + 1 ) e t p ( 0 ) ( k + 1 ) p ( 0 ) log w ( k + 1 )

and

f k + 1 ( k + 1 ) p ( k + 1 ) p ( 0 ) k p ( k ) p ( 0 ) f k g k g k + 1 = f k + 1 ( k + 1 ) p ( k + 1 ) p ( 0 ) g k g k + 1 e t + r p ( 0 ) ,

where e t + r p ( 0 ) > 1 if t < r p ( 0 ) and e t + r p ( 0 ) < 1 if t > r p ( 0 ) , which finishes the proof.□

Corollary 2

Let { f k } k = 1 be an increasing sequence of positive reals. Suppose that the sequence f k k k = 1 is also increasing and there exist r , s , w such that

f k k e r s k s log w k .

Then, we have

(3) lim k f k k k k s log w k = e r 2 s .

Remark 1

Note that identity (2) can be n-times iterated for all n if we make additional assumptions about monotonicity of the corresponding sequences.

3 Applications

Some applications of Theorem 1 are given as follows.

(3.1) Let f k = k . Then, lim k k ! k k = 1 e .

(3.2) Let f k = 1 i = 1 k 1 1 p i . Then, f k e γ log k (see [2,3]) and

lim k f k k log k = lim k i = 1 k 1 1 p i k + i 1 k log k = e γ ,

where γ denotes Euler’s constant. Let α > 0 . If we replace f k by

f k ( α ) = 1 i = i 0 k 1 α p i

for k i 0 , where i 0 min { i : p i > α } , then f k ( α ) e log ( c ( α ) ) log α k for some c ( α ) > 0 (see [2,4,5]), and consequently

lim k f k ( α ) k log α k = 1 c ( α ) .

For example, c ( 2 ) 0.832429065662 .

(3.3) Let α , 0 α < 1 . Then (see [6])

f k = e k k i = 1 k ( i α ) 2 π Γ ( 1 α ) k 1 2 α ,

which implies the relation

lim k f k k k 1 2 α = 2 π Γ ( 1 α ) e α 1 2 .

(3.4) Let us set f n = i = 1 n i n . Then (see [6,14,15]), we have f n e e 1 n n , which implies (see also final remark 2)

lim n f n n n n = e e 1 ,

i.e.,

lim n f n n ( n + 1 ) n = 1 e 1 .

(3.5) Next, we set f n = k = 1 n k k n 2 2 . By the definition of the Glaischer-Kinkelin constant (see [6,7,8])

A = lim n k = 1 n k k n n 2 2 + n 2 + 1 12 e n 2 4 ,

we also have

A = 2 5 36 π 1 6 exp 2 3 0 1 2 log Γ ( t ) d t = exp 1 12 ζ ( 1 ) 1.2824271291 ,

(see [7,9,10,11]) we obtain f n e 1 2 n and

f n k = 1 n k k i = k n 1 i 2 2 ,

which by Theorem 1 gives us the relation

lim n f n n n = e 3 2 .

(3.6) At last, if we set

f n = e n 2 4 k = 1 n k k ,

then f n A n n ( n + 1 ) 2 + 1 12 , where A denotes the Glaisher-Kinkelin constant. Therefore, we get

lim n f n n n ( n + 1 ) 2 n n 12 = A e 1 12 .

Now we present an application of Corollary 2.

(3.7) In [7, Problem 1.5], it was proved that

(4) f k = e k 2 i = 1 k k i k e 1 ln 2 π k 1 2 .

Hence, by Theorem 1, we get

lim k k f k k = e 3 2 ln 2 π

and by Corollary 2

lim k k f k k k = e 2 ln 2 π .

We note that from (4) we obtain the solution of Problem 51, p. 45 from Pólya and Szegő [12]:

lim k i = 1 k k i k 2 = e .

Note also that, applying Corollary 2, monotonicity of sequences { f k } k = 1 and f k k k = 1 is not important at all, because the following lemma holds.

Lemma 2

Let { a k } k = 1 be a sequence of positive reals. If lim k a k = α , then there exists an increasing sequence { A k } k = 1 such that lim k A k = α and lim k A k a k = 1 . We may assume that the sequence ( A k ) k k = 1 is increasing as well.

(3.8) Also from [7, Problem 1.14] we get

lim k Γ 1 k k k = e 1 ,

which by Corollary 2 implies

lim k 1 k Γ 1 k k k = e 2 .

4 Final remarks

  1. Using the other asymptotic expansions, especially the ones given in the study of Kellner [6] (e.g., for product of Bernoulli numbers, for products of the special values of gamma function, etc.), we can generate many new relations that are omitted here. Other relations were published recently in [13] as well.

  2. In the history of the following asymptotic relation (see [14,15,16])

n n k = 1 n k n = e e 1 e ( e + 1 ) 2 n ( e 1 ) 3 + o ( n 1 ) ,

one thread – Dutch connection is missing. The above relationship was found by the last author as one of the problems in the problem section of the known Dutch journal Nieuw Archief voor Wiskunde but in the equivalent form:

( n + 1 ) n k = 1 n k n = 1 e 1 e ( 3 e ) 2 n ( e 1 ) 3 + o ( n 1 ) .

It is interesting that in both proofs of these equalities (from Nieuw Archief voor Wiskunde and by Lampret – see [15]) Tannery’s theorem was used.

  1. Another estimation for a number of primes is discussed by Meštrović in [17].

5 Problems

The natural question about asymptoticity of the following expressions:

f k k p ( k ) p ( 0 ) k k p ( 0 ) log w k e r p ( 0 )

and

f k k k k s log w k e r 2 s

arises from (1) and (2). For now, this question remains unanswered. The following formula (see [18,19])

e ( 1 ) n n e 1 4 e 1 3 e 1 2 e 1 ( 1 + x ) 1 x 1 x 1 x 1 x 1 x 1 x 1 x x 0 e ( 1 ) n n + 1

is an inspiration for solving this problem.

Acknowledgments

The authors would like to thank reviewers for invaluable comments which help to improve the presentation of this study.

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Received: 2019-12-02
Revised: 2020-06-03
Accepted: 2020-06-12
Published Online: 2020-08-04

© 2020 Agata Chmielowska et al., published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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