New upper bounds of n!

  • Mansour Mahmoud1, 3Email author,

    Affiliated with

    • Mohammed A Alghamdi1 and

      Affiliated with

      • Ravi P Agarwal2

        Affiliated with

        Journal of Inequalities and Applications20122012:27

        DOI: 10.1186/1029-242X-2012-27

        Received: 15 October 2011

        Accepted: 13 February 2012

        Published: 13 February 2012

        Abstract

        In this article, we deduce a new family of upper bounds of n! of the form

        n ! < 2 π n ( n / e ) n e M n [ m ] n , M n [ m ] = 1 2 m + 3 1 4 n + k = 1 m 2 m - 2 k + 2 2 k + 1 2 - 2 k ζ ( 2 k , n + 1 / 2 ) m = 1 , 2 , 3 , . . . . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equa_HTML.gif

        We also proved that the approximation formula 2 π n ( n / e ) n e M n [ m ] http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_IEq1_HTML.gif for big factorials has a speed of convergence equal to n-2m- 3for m = 1,2,3,..., which give us a superiority over other known formulas by a suitable choice of m.

        Mathematics Subject Classification (2000): 41A60; 41A25; 57Q55; 33B15; 26D07.

        Keywords

        Stirling' formula Wallis' formula Bernoulli numbers Riemann Zeta function speed of convergence

        1 Introduction

        Stirling' formula
        n ! ~ 2 n π ( n / e ) n http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equ1_HTML.gif
        (1)
        is one of the most widely known and used in asymptotics. In other words, we have
        lim n n ! e n 2 π n n n = 1 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equ2_HTML.gif
        (2)
        This formula provides an extremely accurate approximation of n! for large values of n. The first proofs of Stirling's formula was given by De Moivre (1730) [1] and Stirling (1730) [2]. Both used what is now called the Euler-MacLaurin formula to approximate log 2 + log 3 + ... + log n. The first derivation of De Moivre did not explicity determine the constant 2 π http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_IEq2_HTML.gif. In 1731, Stirling determine this constant using Wallis' formula
        lim n 2 2 n ( n ! ) 2 ( 2 n ) ! 1 n = π . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equb_HTML.gif
        Over the years, there have been many different upper and lower bounds for n! by various authors [310]. Artin [11] show that μ ( n ) = ln n ! e n n n 2 π n http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_IEq3_HTML.gif lies between any two successive partial sums of the Stirling's series
        B 2 1 . 2 . n + B 4 3 . 4 . n 3 + B 6 5 . 6 . n 5 + , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equ3_HTML.gif
        (3)
        where the numbers B i are called the Bernoulli numbers and are defined by
        B 0 = 1 , k = 0 n 1 ( k n ) B k = 0 , n 2. http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equ4_HTML.gif
        (4)

        We can't take infinite sum of the series (3) because the series diverges. Also, Impens [12] deduce Artin result with different proof and show that the Bernoulli numbers in this series cannot be improved by any method whatsoever.

        The organization of this article is as follows. In Section 2, we deduce a general double inequality of n!, which already obtained in [9] with different proof. Section 3 is devoted to getting a new family of upper bounds of n! different from the partial sums of the Stirling's series. In Section 4, we measure the speed of convergence of our approximation formula 2 π n ( n / e ) n e M n [ m ] http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_IEq4_HTML.gif for big factorials. Also, we offer some numerical computations to prove the superiority of our formula over other known formulas.

        2 A double inequality of n!

        In view of the relation (2), we begin with the two sequences K n and f n defined by
        f n = n ! e n 2 π n n n e - K n n 1 , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equ5_HTML.gif
        (5)
        where
        lim n K n = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equ6_HTML.gif
        (6)
        Then we have
        lim n f n = 1 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equ7_HTML.gif
        (7)
        Now define the sequence g n by
        g n = f n + 1 f n = e K n - K n + 1 + 1 1 + 1 n - n - 1 / 2 , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equ8_HTML.gif
        (8)
        which satisfies
        lim n g n = 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equ9_HTML.gif
        (9)
        and
        g n = g n d d n K n - K n + 1 - ( n + 1 / 2 ) ln 1 + 1 n . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equ10_HTML.gif
        (10)
        There are two cases. The first case if K n = a n such that
        d d n a n - a n + 1 - ( n + 1 / 2 ) ln 1 + 1 n > 0 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equ11_HTML.gif
        (11)
        then we get g n > 0 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_IEq5_HTML.gif and hence g n is strictly increasing function. But g n → 1 as n → ∞, then g n < 1. Hence f n + 1 f n < 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_IEq6_HTML.gif which give us that fn+1< f n . Then f n is strictly decreasing function. Also, f n → 1 as n → ∞, then we obtain f n > 1. Then
        n ! e n 2 π n n n e - a n > 1 n 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equ12_HTML.gif
        (12)
        or
        2 π n ( n / e ) n e a n < n ! n 1 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equ13_HTML.gif
        (13)
        The condition (11) means that the function a n - an+1-(n+ 1/2) ln 1 + 1 n http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_IEq7_HTML.gif is strictly increasing function also it tends to -1 as n → ∞. Then
        a n - a n + 1 - ( n + 1 / 2 ) ln 1 + 1 n < - 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equc_HTML.gif
        or
        a n - a n + 1 < ( n + 1 / 2 ) ln 1 + 1 n - 1 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equ14_HTML.gif
        (14)
        The second case if K n = b n such that
        d d n b n - b n + 1 - ( n + 1 / 2 ) ln 1 + 1 n < 0 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equ15_HTML.gif
        (15)
        Similarly, we can prove that f n < 1. Then
        n ! e n 2 π n n n e - b n < 1 n 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equ16_HTML.gif
        (16)
        or
        n ! < 2 π n ( n / e ) n e b n n 1 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equ17_HTML.gif
        (17)
        The condition (15) means that the function b n - bn+1- (n + 1/2) ln 1 + 1 n http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_IEq7_HTML.gif is strictly decreasing function also it tends to -1 as n → ∞. Then
        b n - b n + 1 - ( n + 1 / 2 ) ln 1 + 1 n > - 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equd_HTML.gif
        or
        ( n + 1 / 2 ) ln 1 + 1 n - 1 < b n - b n + 1 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equ18_HTML.gif
        (18)
        From the well-known expansion
        1 2 ln 1 + y 1 - y = k = 1 y 2 k - 1 2 k - 1 y < 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Eque_HTML.gif
        in which we substitute y = 1 2 n + 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_IEq8_HTML.gif, so 1 + y 1 - y = 1 + 1 n http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_IEq9_HTML.gif. Then
        ( n + 1 / 2 ) ln 1 + 1 n - 1 = k = 1 1 2 k + 1 1 ( 2 n + 1 ) 2 k . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equ19_HTML.gif
        (19)

        Now we obtain the following result

        Theorem 1.
        2 π n ( n / e ) n e a n < n ! < 2 π n ( n / e ) n e b n n 1 , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equ20_HTML.gif
        (20)
        where the two sequences a n , b n → 0 as n → ∞ and satisfy
        a n - a n + 1 < k = 1 1 2 k + 1 1 ( 2 n + 1 ) 2 k < b n - b n + 1 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equ21_HTML.gif
        (21)

        A q-analog of the inequality (20) was introduced in [13].

        3 A new family of upper bounds of n!

        By manipulating the series k = 1 1 2 k + 1 1 ( 2 n + 1 ) 2 k http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_IEq10_HTML.gif to find upper bounds, we get
        k = 1 1 2 k + 1 1 ( 2 n + 1 ) 2 k < k = 1 m 1 ( 2 k + 1 ) ( 2 n + 1 ) 2 k + 1 2 m + 3 k = m + 1 1 ( 2 n + 1 ) 2 k < k = 1 1 ( 2 k + 1 ) ( 2 n + 1 ) 2 k + 1 4 ( 2 m + 3 ) n ( n + 1 ) ( 2 n + 1 ) 2 m , m = 1 , 2 , 3 , . . . . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equf_HTML.gif
        Let's solve the following recurrence relation w.r.t n
        M n [ m ] - M n + 1 [ m ] = k = 1 m 1 ( 2 k + 1 ) ( 2 n + 1 ) 2 k + 1 4 ( 2 m + 3 ) n ( n + 1 ) ( 2 n + 1 ) 2 m , m = 1 , 2 , 3 , . . . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equ22_HTML.gif
        (22)
        which give us
        M n [ m ] = M 0 [ m ] - i = 1 n - 1 k = 1 m 1 ( 2 k + 1 ) ( 2 i + 1 ) 2 k + 1 4 ( 2 m + 3 ) i ( i + 1 ) ( 2 i + 1 ) 2 m = M 0 [ m ] - k = 1 m 1 2 k + 1 i = 1 n - 1 1 ( 2 i + 1 ) 2 k - 1 4 ( 2 m + 3 ) i = 1 n - 1 1 i ( i + 1 ) ( 2 i + 1 ) 2 m . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equg_HTML.gif
        But
        i = 1 n - 1 1 i ( i + 1 ) ( 2 i + 1 ) 2 m = i = 1 n - 1 1 i - 1 i + 1 - 4 ( 2 i + 1 ) 2 - 4 ( 2 i + 1 ) 4 - - 4 ( 2 i + 1 ) 2 m = n - 1 n - 4 k = 1 m i = 1 n - 1 1 ( 2 i + 1 ) 2 k . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equh_HTML.gif
        Then
        M n [ m ] = M 0 [ m ] - n - 1 4 n ( 2 m + 3 ) - k = 1 m 1 2 k + 1 - 1 2 m + 3 i = 1 n - 1 1 ( 2 i + 1 ) 2 k . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equi_HTML.gif
        The series
        i = 1 1 ( 2 i + 1 ) 2 k = ζ ( 2 k ) ( 1 - 2 - 2 k ) - 1 , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equj_HTML.gif
        where ζ(x) is the Riemann Zeta function. By using the relation [14]
        ζ ( 2 k ) = ( - 1 ) k - 1 2 2 k - 1 ( 2 k ) ! B 2 k π 2 k , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equk_HTML.gif
        where B r s http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_IEq11_HTML.gif are Bernoulli's numbers. Then
        i = 1 1 ( 2 i + 1 ) 2 k = ( - 1 ) k - 1 ( 2 2 k - 1 ) 2 ( 2 k ) ! B 2 k π 2 k - 1 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equl_HTML.gif
        Hence, we can choose
        M 0 [ m ] = 1 4 ( 2 m + 3 ) + k = 1 m 1 2 k + 1 - 1 2 m + 3 ( - 1 ) k - 1 ( 2 2 k - 1 ) 2 ( 2 k ) ! B 2 k π 2 k - 1 , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equ23_HTML.gif
        (23)
        which satisfies
        lim n M n [ m ] = 0 , m = 1 , 2 , 3 , . . . . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equm_HTML.gif
        Then
        M n [ m ] = 1 2 m + 3 1 4 n + k = 1 m 2 m - 2 k + 2 2 k + 1 ( - 1 ) k - 1 ( 2 2 k - 1 ) 2 ( 2 k ) ! B 2 k π 2 k - 1 - i = 1 n - 1 1 ( 2 i + 1 ) 2 k . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equ24_HTML.gif
        (24)
        By using the relation
        i = 1 n - 1 1 ( 2 i + 1 ) 2 k = - 1 - ( 2 - 2 k - 1 ) ζ ( 2 k ) - 2 - 2 k ζ ( 2 k , n + 1 / 2 ) = - 1 - ( - 1 ) k - 1 ( 1 - 2 2 k ) 2 ( 2 k ) ! B 2 k π 2 k - 2 - 2 k ζ ( 2 k , n + 1 / 2 ) , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equn_HTML.gif
        we get
        M n [ m ] = 1 2 m + 3 1 4 n + k = 1 m 2 m - 2 k + 2 2 k + 1 2 - 2 k ζ ( 2 k , n + 1 / 2 ) . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equ25_HTML.gif
        (25)

        In the following Lemma, we will see that the upper bound M n m http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_IEq12_HTML.gif will improved with increasing the value of m.

        Lemma 3.1.
        M n [ m + 1 ] < M n [ m ] , m , n = 1 , 2 , 3 , . . . . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equ26_HTML.gif
        (26)
        Proof. From (25), we get
        M n [ m + 1 ] = 1 2 m + 5 1 4 n + k = 1 m 2 m - 2 k + 2 2 k + 1 2 - 2 k ζ ( 2 k , n + 1 / 2 ) + 1 2 m + 5 k = 1 m + 1 1 2 k + 1 2 1 - 2 k ζ ( 2 k , n + 1 / 2 ) . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equo_HTML.gif
        Then
        M n [ m + 1 ] - M n [ m ] = 2 ( 2 m + 3 ) ( 2 m + 5 ) - 1 4 n + k = 1 m + 1 2 - 2 k ζ ( 2 k , n + 1 / 2 ) < 2 ( 2 m + 3 ) ( 2 m + 5 ) - 1 4 n + k = 1 2 - 2 k ζ ( 2 k , n + 1 / 2 ) . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equp_HTML.gif
        Now
        k = 1 2 - 2 k ζ ( 2 k , n + 1 / 2 ) = k = 1 r = 0 2 - 2 k ( n + r + 1 / 2 ) 2 k = r = 0 k = 1 2 - 2 k ( n + r + 1 / 2 ) 2 k = r = 0 1 1 - 1 ( 2 n + 2 r + 1 ) 2 - 1 = r = 0 1 4 ( n + r ) ( n + r + 1 ) = 1 4 r = 0 1 n + r - 1 n + r + 1 = 1 4 n . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equq_HTML.gif
        Then
        k = 1 2 - 2 k ζ ( 2 k , n + 1 / 2 ) = 1 4 n . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equ27_HTML.gif
        (27)
        and hence
        M n [ m + 1 ] - M n [ m ] < 2 ( 2 m + 3 ) ( 2 m + 5 ) - 1 4 n + k = 1 2 - 2 k ζ ( 2 k , n + 1 / 2 ) < 2 ( 2 m + 3 ) ( 2 m + 5 ) - 1 4 n + 1 4 n < 0 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equr_HTML.gif

        The following Lemma show that M n [ 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_IEq13_HTML.gif is an improvement of the upper bound 1 12 n http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_IEq14_HTML.gif.

        Lemma 3.2.
        M n [ 1 ] < 1 12 n http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equ28_HTML.gif
        (28)
        Proof. From Eq. (25), we have M n [ 1 ] = 1 20 n + 1 30 ζ ( 2 , n + 1 / 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_IEq15_HTML.gif. By using (27), we obtain
        1 n = k = 1 2 2 - 2 k ζ ( 2 k , n + 1 / 2 ) = ζ ( 2 , n + 1 / 2 ) + k = 2 2 2 - 2 k ζ ( 2 k , n + 1 / 2 ) . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equs_HTML.gif
        Then
        ζ ( 2 , n + 1 / 2 ) < 1 n http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equt_HTML.gif
        which give us that
        1 30 ζ ( 2 , n + 1 / 2 ) + 1 20 n < 1 30 n + 1 20 n = 1 12 n . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equu_HTML.gif

        4 The speed of convergence of the approximation formula 2 π n ( n / e ) n e M n [ m ] http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_IEq16_HTML.gif for big factorials

        In what follows, we need the following result, which represents a powerful tool to measure the rate of convergence.

        Lemma 4.1. If (w n )n≥1is convergent to zero and there exists the limit
        lim n n k ( w n - w n + 1 ) = l http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equ29_HTML.gif
        (29)
        with k > 1, then there exists the limit:
        lim n n k - 1 w n = l k - 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equv_HTML.gif

        This Lemma was first used by Mortici for constructing asymptotic expansions, or to accelerate some convergences [1521]. By using Lemma (4.1), clearly the sequence (w n )n≥1converges more quickly when the value of k satisfying (29) is larger.

        To measure the accuracy of approximation formula 2 π n ( n / e ) n e M n [ m ] http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_IEq16_HTML.gif, define the sequence (w n )n≥1by the relation
        n ! = 2 π n ( n / e ) n e M n [ m ] e w n ; n = 1 , 2 , 3 , . . . . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equ30_HTML.gif
        (30)
        This approximation formula will be better as (w n )n≥1converges faster to zero. Using the relation (30), we get
        w n = ln n ! - ln 2 π - ( n + 1 / 2 ) ln n + n - M n [ m ] http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equw_HTML.gif
        Then
        w n - w n + 1 = ( n + 1 / 2 ) ln ( 1 + 1 / n ) - 1 + M n + 1 [ m ] - M n [ m ] http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equx_HTML.gif
        By using the relations (19) and (22), we have
        w n - w n + 1 = k = m + 1 1 2 k + 1 1 ( 2 n + 1 ) 2 k - 1 4 ( 2 m + 3 ) n ( n + 1 ) ( 2 n + 1 ) 2 m = k = m + 3 1 2 k + 1 1 ( 2 n + 1 ) 2 k - ( 5 + 2 m + 8 n + 8 n 2 ) ( 2 n + 1 ) - 4 - 2 m 4 n ( n + 1 ) ( 15 + 16 m + 4 m 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equy_HTML.gif
        Then
        lim n n 2 ( m + 2 ) ( w n - w n + 1 ) = - 1 ( 2 m + 3 ) ( 2 m + 5 ) 2 2 m + 3 ; n , m = 1 , 2 , 3 , . . . . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equ31_HTML.gif
        (31)
        Theorem 2. The rate of convergence of the sequence w n is equal to n-2m-3, since
        lim n n 2 m + 3 w n = 1 ( 2 m + 3 ) 2 ( 2 m + 5 ) 2 2 m + 3 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equz_HTML.gif
        In 2011, Mortici [22] shows by numerical computations that his formula
        n ! ~ 2 π n ( n / e ) n e 1 12 n + 2 5 n = μ n http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_Equ32_HTML.gif
        (32)
        is much stronger than other known formulas such as:1

        n ! ~ 2 π n + 1 / 2 e n + 1 / 2 = β n http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_IEq17_HTML.gif

        (Burnside [23])

        n ! ~ 2 π e - n n n + 1 n - 1 / 6 = δ n http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_IEq18_HTML.gif

        (Batir [24])

        n ! ~ ( 2 n + 1 / 3 ) π ( n / e ) n = γ n http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_IEq19_HTML.gif

        (Gosper [25])

        n ! ~ π 8 n 3 + 4 n 2 + n + 1 / 30 6 ( n / e ) n = ρ n http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_IEq20_HTML.gif

        (Ramanujan [26])

        The following table shows numerically that our new formula λ n , 1 = 2 π n ( n / e ) n e M n [ 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-27/MediaObjects/13660_2011_Article_181_IEq21_HTML.gif has a superiority over the the Mortici's formula μ n .

        n

        |n! -μ n |

        |n! - λn,1|

        10

        0.0252281

        0.00641793

        25

        1.1 × 1015

        2.8 × 1014

        50

        6.8 × 1052

        1.7 × 1052

        100

        6.5 × 10144

        1.4 × 10144

        Declarations

        Acknowledgements

        This project was founded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah, under grant No. 448/130/1431. The authors, therefore, acknowledge with thanks DSR support for Scientific Research.

        Authors’ Affiliations

        (1)
        Mathematics Department, Faculty of Science, King Abdulaziz University
        (2)
        Department of Mathematics, Texas A&M University
        (3)
        Department of Mathematics, Faculty of Science, Mansoura University

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        © Mahmoud et al; licensee Springer. 2012

        This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://​creativecommons.​org/​licenses/​by/​2.​0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.