New upper bounds of n!

Abstract

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

$$\begin{array}{c}n!<\sqrt{2\pi n}{\left(n/e\right)}^n{e}^{M_n^{\left[m\right]}}n\in \mathbb{N},\\ M_n^{\left[m\right]}=\frac{1}{2m+3}\left[\frac{1}{4n}+\sum _{k=1}^m\frac{2m-2k+2}{2k+1}{2}^{-2k}\zeta \left(2k,n+1/2\right)\right]m=1,2,3,....\end{array}$$

We also proved that the approximation formula $\sqrt{2\pi n}{\left(n/e\right)}^n{e}^{M_n^{\left[m\right]}}$ for big factorials has a speed of convergence equal to n^{-2m- 3}for 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.

1 Introduction

Stirling' formula

$$n!~\sqrt{2n\pi }{\left(n/e\right)}^n$$

(1)

is one of the most widely known and used in asymptotics. In other words, we have

$$\underset{n\to \infty }{\text{lim}}\frac{n!e^n}{\sqrt{2\pi n}{n}^n}=1.$$

(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 $\sqrt{2\pi }$. In 1731, Stirling determine this constant using Wallis' formula

$$\underset{n\to \infty }{\text{lim}}\frac{2^{2n}{\left(n!\right)}^2}{\left(2n\right)!}\frac{1}{\sqrt{n}}=\sqrt{\pi }.$$

Over the years, there have been many different upper and lower bounds for n! by various authors [3–10]. Artin [11] show that $\mu \left(n\right)=\text{ln}\frac{n!e^n}{n^n\sqrt{2\pi n}}$ lies between any two successive partial sums of the Stirling's series

$$\frac{B_2}{1.2.n}+\frac{B_4}{3.4.n^3}+\frac{B_6}{5.6.n^5}+\cdots ,$$

(3)

where the numbers B _i are called the Bernoulli numbers and are defined by

$$B_0=1,{\displaystyle \sum _{k=0}^{n-1}{(}_k^n)}{B}_k=0,n\ge 2.$$

(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 $\sqrt{2\pi n}{\left(n/e\right)}^n{e}^{M_n^{\left[m\right]}}$ 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

$$f_n=\frac{n!e^n}{\sqrt{2\pi n}{n}^n}{e}^{-K_n}n\ge 1,$$

(5)

where

$$\underset{n\to \infty }{\text{lim}}{K}_n=0.$$

(6)

Then we have

$$\underset{n\to \infty }{\text{lim}}{f}_n=1.$$

(7)

Now define the sequence g _n by

$$g_n=\frac{f_{n+1}}{f_n}=e^{K_n-K_{n+1}+1}{\left(1+\frac{1}{n}\right)}^{-n-1/2},$$

(8)

which satisfies

$$\underset{n\to \infty }{\text{lim}}{g}_n=1$$

(9)

and

$$g_n^{\prime }=g_n\frac{d}{dn}\left(K_n-K_{n+1}-\left(n+1/2\right)\text{ln}\left(1+\frac{1}{n}\right)\right).$$

(10)

There are two cases. The first case if K _n = a _n such that

$$\frac{d}{dn}\left(a_n-a_{n+1}-\left(n+1/2\right)\text{ln}\left(1+\frac{1}{n}\right)\right)>0$$

(11)

then we get $g_n^{\prime }>0$ and hence g _n is strictly increasing function. But g _n → 1 as n → ∞, then g _n < 1. Hence $\frac{f_{n+1}}{f_n}<1$ which give us that f_n+1< f _n . Then f _n is strictly decreasing function. Also, f _n → 1 as n → ∞, then we obtain f _n > 1. Then

$$\frac{n!e^n}{\sqrt{2\pi n}{n}^n}{e}^{-a_n}>1n\ge 1$$

(12)

or

$$\sqrt{2\pi n}{\left(n/e\right)}^n{e}^{a_n}<n!n\ge 1.$$

(13)

The condition (11) means that the function a _n - a_n+1-(n+ 1/2) ln $\left(1+\frac{1}{n}\right)$ is strictly increasing function also it tends to -1 as n → ∞. Then

$$a_n-a_{n+1}-\left(n+1/2\right)\text{ln}\left(1+\frac{1}{n}\right)<-1$$

or

$$a_n-a_{n+1}<\left(n+1/2\right)\text{ln}\left(1+\frac{1}{n}\right)-1.$$

(14)

The second case if K _n = b _n such that

$$\frac{d}{dn}\left(b_n-b_{n+1}-\left(n+1/2\right)\text{ln}\left(1+\frac{1}{n}\right)\right)<0.$$

(15)

Similarly, we can prove that f _n < 1. Then

$$\frac{n!e^n}{\sqrt{2\pi n}{n}^n}{e}^{-b_n}<1n\ge 1$$

(16)

or

$$n!<\sqrt{2\pi n}{\left(n/e\right)}^n{e}^{b_n}n\ge 1.$$

(17)

The condition (15) means that the function b _n - b_n+1- (n + 1/2) ln $\left(1+\frac{1}{n}\right)$ is strictly decreasing function also it tends to -1 as n → ∞. Then

$$b_n-b_{n+1}-\left(n+1/2\right)\text{ln}\left(1+\frac{1}{n}\right)>-1$$

or

$$\left(n+1/2\right)\text{ln}\left(1+\frac{1}{n}\right)-1<b_n-b_{n+1}.$$

(18)

From the well-known expansion

$$\frac{1}{2}\text{ln}\left(\frac{1+y}{1-y}\right)=\sum _{k=1}^{\infty }\frac{y^{2k-1}}{2k-1}\left|y\right|<1$$

in which we substitute $y=\frac{1}{2n+1}$, so $\frac{1+y}{1-y}=1+\frac{1}{n}$. Then

$$\left(n+1/2\right)\text{ln}\left(1+\frac{1}{n}\right)-1=\sum _{k=1}^{\infty }\frac{1}{2k+1}\frac{1}{{\left(2n+1\right)}^{2k}}.$$

(19)

Now we obtain the following result

Theorem 1.

$$\sqrt{2\pi n}{\left(n/e\right)}^n{e}^{a_n}<n!<\sqrt{2\pi n}{\left(n/e\right)}^n{e}^{b_n}n\ge 1,$$

(20)

where the two sequences a _n , b _n → 0 as n → ∞ and satisfy

$$a_n-a_{n+1}<\sum _{k=1}^{\infty }\frac{1}{2k+1}\frac{1}{{\left(2n+1\right)}^{2k}}<b_n-b_{n+1}.$$

(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 ${\sum }_{k=1}^{\infty }\frac{1}{2k+1}\frac{1}{{\left(2n+1\right)}^{2k}}$ to find upper bounds, we get

$$\begin{array}{c}\sum _{k=1}^{\infty }\frac{1}{2k+1}\frac{1}{{\left(2n+1\right)}^{2k}}\\ <\sum _{k=1}^m\frac{1}{\left(2k+1\right){\left(2n+1\right)}^{2k}}+\frac{1}{2m+3}\sum _{k=m+1}^{\infty }\frac{1}{{\left(2n+1\right)}^{2k}}\\ <\sum _{k=1}^{\infty }\frac{1}{\left(2k+1\right){\left(2n+1\right)}^{2k}}+\frac{1}{4\left(2m+3\right)n\left(n+1\right){\left(2n+1\right)}^{2m}},m=1,2,3,....\end{array}$$

Let's solve the following recurrence relation w.r.t n

$$M_n^{\left[m\right]}-M_{n+1}^{\left[m\right]}=\sum _{k=1}^m\frac{1}{\left(2k+1\right){\left(2n+1\right)}^{2k}}+\frac{1}{4\left(2m+3\right)n\left(n+1\right){\left(2n+1\right)}^{2m}},m=1,2,3,...$$

(22)

which give us

$$\begin{array}{ll}\hfill M_n^{\left[m\right]}& =M_0^{\left[m\right]}-\sum _{i=1}^{n-1}\left(\sum _{k=1}^m\frac{1}{\left(2k+1\right){\left(2i+1\right)}^{2k}}+\frac{1}{4\left(2m+3\right)i\left(i+1\right){\left(2i+1\right)}^{2m}}\right)\\ =M_0^{\left[m\right]}-\sum _{k=1}^m\frac{1}{2k+1}\left(\sum _{i=1}^{n-1}\frac{1}{{\left(2i+1\right)}^{2k}}\right)-\frac{1}{4\left(2m+3\right)}\sum _{i=1}^{n-1}\frac{1}{i\left(i+1\right){\left(2i+1\right)}^{2m}}.\end{array}$$

But

$$\begin{array}{ll}\hfill \sum _{i=1}^{n-1}\frac{1}{i\left(i+1\right){\left(2i+1\right)}^{2m}}& =\sum _{i=1}^{n-1}\left(\frac{1}{i}-\frac{1}{i+1}-\frac{4}{{\left(2i+1\right)}^2}-\frac{4}{{\left(2i+1\right)}^4}-\cdots -\frac{4}{{\left(2i+1\right)}^{2m}}\right)\\ =\frac{n-1}{n}-4\sum _{k=1}^m\sum _{i=1}^{n-1}\frac{1}{{\left(2i+1\right)}^{2k}}.\end{array}$$

Then

$$M_n^{\left[m\right]}=M_0^{\left[m\right]}-\frac{n-1}{4n\left(2m+3\right)}-\sum _{k=1}^m\left(\frac{1}{2k+1}-\frac{1}{2m+3}\right)\left(\sum _{i=1}^{n-1}\frac{1}{{\left(2i+1\right)}^{2k}}\right).$$

The series

$$\sum _{i=1}^{\infty }\frac{1}{{\left(2i+1\right)}^{2k}}=\zeta \left(2k\right)\left(1-2^{-2k}\right)-1,$$

where ζ(x) is the Riemann Zeta function. By using the relation [14]

$$\zeta \left(2k\right)=\frac{{\left(-1\right)}^{k-1}{2}^{2k-1}}{\left(2k\right)!}{B}_{2k}{\pi }^{2k},$$

where $B_r^{\prime }s$ are Bernoulli's numbers. Then

$$\sum _{i=1}^{\infty }\frac{1}{{\left(2i+1\right)}^{2k}}=\frac{{\left(-1\right)}^{k-1}\left(2^{2k}-1\right)}{2\left(2k\right)!}{B}_{2k}{\pi }^{2k}-1.$$

Hence, we can choose

$$M_0^{\left[m\right]}=\frac{1}{4\left(2m+3\right)}+\sum _{k=1}^m\left(\frac{1}{2k+1}-\frac{1}{2m+3}\right)\left(\frac{{\left(-1\right)}^{k-1}\left(2^{2k}-1\right)}{2\left(2k\right)!}{B}_{2k}{\pi }^{2k}-1\right),$$

(23)

which satisfies

$$\underset{n\to \infty }{\text{lim}}{M}_n^{\left[m\right]}=0,m=1,2,3,....$$

Then

$$M_n^{\left[m\right]}=\frac{1}{2m+3}\left[\frac{1}{4n}+\sum _{k=1}^m\frac{2m-2k+2}{2k+1}\left(\frac{{\left(-1\right)}^{k-1}\left(2^{2k}-1\right)}{2\left(2k\right)!}{B}_{2k}{\pi }^{2k}-1-\sum _{i=1}^{n-1}\frac{1}{{\left(2i+1\right)}^{2k}}\right)\right].$$

(24)

By using the relation

$$\begin{array}{ll}\hfill \sum _{i=1}^{n-1}\frac{1}{{\left(2i+1\right)}^{2k}}& =-1-\left(2{-}^{2k}-1\right)\zeta \left(2k\right)-2^{-2k}\zeta \left(2k,n+1/2\right)\\ =-1-\frac{{\left(-1\right)}^{k-1}\left(1-2^{2k}\right)}{2\left(2k\right)!}{B}_{2k}{\pi }^{2k}-2^{-2k}\zeta \left(2k,n+1/2\right),\end{array}$$

we get

$$M_n^{\left[m\right]}=\frac{1}{2m+3}\left[\frac{1}{4n}+\sum _{k=1}^m\frac{2m-2k+2}{2k+1}{2}^{-2k}\zeta \left(2k,n+1/2\right)\right].$$

(25)

In the following Lemma, we will see that the upper bound $M_n^m$ will improved with increasing the value of m.

Lemma 3.1.

$$M_n^{\left[m+1\right]}<M_n^{\left[m\right]},m,n=1,2,3,....$$

(26)

Proof. From (25), we get

$$\begin{array}{ll}\hfill M_n^{\left[m+1\right]}& =\frac{1}{2m+5}\left[\frac{1}{4n}+\sum _{k=1}^m\frac{2m-2k+2}{2k+1}{2}^{-2k}\zeta \left(2k,n+1/2\right)\right]\\ +\frac{1}{2m+5}\sum _{k=1}^{m+1}\frac{1}{2k+1}{2}^{1-2k}\zeta \left(2k,n+1/2\right).\end{array}$$

Then

$$\begin{array}{ll}\hfill M_n^{\left[m+1\right]}-M_n^{\left[m\right]}& =\frac{2}{\left(2m+3\right)\left(2m+5\right)}\left[\frac{-1}{4n}+\sum _{k=1}^{m+1}{2}^{-2k}\zeta \left(2k,n+1/2\right)\right]\\ <\frac{2}{\left(2m+3\right)\left(2m+5\right)}\left[\frac{-1}{4n}+\sum _{k=1}^{\infty }{2}^{-2k}\zeta \left(2k,n+1/2\right)\right].\end{array}$$

Now

$$\begin{array}{ll}\hfill \sum _{k=1}^{\infty }{2}^{-2k}\zeta \left(2k,n+1/2\right)& =\sum _{k=1}^{\infty }\sum _{r=0}^{\infty }\frac{2^{-2k}}{{\left(n+r+1/2\right)}^{2k}}\\ =\sum _{r=0}^{\infty }\sum _{k=1}^{\infty }\frac{2^{-2k}}{{\left(n+r+1/2\right)}^{2k}}\\ =\sum _{r=0}^{\infty }\left[\frac{1}{1-\left(\frac{1}{{\left(2n+2r+1\right)}^2}\right)}-1\right]\\ =\sum _{r=0}^{\infty }\frac{1}{4\left(n+r\right)\left(n+r+1\right)}\\ =\frac{1}{4}\sum _{r=0}^{\infty }\left[\frac{1}{n+r}-\frac{1}{n+r+1}\right]\\ =\frac{1}{4n}.\end{array}$$

Then

$$\sum _{k=1}^{\infty }{2}^{-2k}\zeta \left(2k,n+1/2\right)=\frac{1}{4n}.$$

(27)

and hence

$$\begin{array}{ll}\hfill M_n^{\left[m+1\right]}-M_n^{\left[m\right]}& <\frac{2}{\left(2m+3\right)\left(2m+5\right)}\left[\frac{-1}{4n}+\sum _{k=1}^{\infty }{2}^{-2k}\zeta \left(2k,n+1/2\right)\right]\\ <\frac{2}{\left(2m+3\right)\left(2m+5\right)}\left[\frac{-1}{4n}+\frac{1}{4n}\right]\\ <0.\end{array}$$

The following Lemma show that $M_n^{\left[1\right]}$ is an improvement of the upper bound $\frac{1}{12n}$.

Lemma 3.2.

$$M_n^{\left[1\right]}<\frac{1}{12n}$$

(28)

Proof. From Eq. (25), we have $M_n^{\left[1\right]}=\frac{1}{20n}+\frac{1}{30}\zeta \left(2,n+1/2\right)$. By using (27), we obtain

$$\begin{array}{ll}\hfill \frac{1}{n}& =\sum _{k=1}^{\infty }{2}^{2-2k}\zeta \left(2k,n+1/2\right)\\ =\zeta \left(2,n+1/2\right)+\sum _{k=2}^{\infty }{2}^{2-2k}\zeta \left(2k,n+1/2\right).\end{array}$$

Then

$$\zeta \left(2,n+1/2\right)<\frac{1}{n}$$

which give us that

$$\frac{1}{30}\zeta \left(2,n+1/2\right)+\frac{1}{20n}<\frac{1}{30n}+\frac{1}{20n}=\frac{1}{12n}.$$

4 The speed of convergence of the approximation formula $\sqrt{2\pi n}{\left(n/e\right)}^n{e}^{M_n^{\left[m\right]}}$ 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

$$\underset{n\to \infty }{\text{lim}}{n}^k\left(w_n-w_{n+1}\right)=l\in ℝ$$

(29)

with k > 1, then there exists the limit:

$$\underset{n\to \infty }{\text{lim}}{n}^{k-1}{w}_n=\frac{l}{k-1}$$

This Lemma was first used by Mortici for constructing asymptotic expansions, or to accelerate some convergences [15–21]. 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 $\sqrt{2\pi n}{\left(n/e\right)}^n{e}^{M_n^{\left[m\right]}}$, define the sequence (w _n )_n≥1by the relation

$$n!=\sqrt{2\pi n}{\left(n/e\right)}^n{e}^{M_n^{\left[m\right]}}{e}^{w_n};n=1,2,3,....$$

(30)

This approximation formula will be better as (w _n )_n≥1converges faster to zero. Using the relation (30), we get

$$w_n=\text{ln}n!-\text{ln}\sqrt{2\pi }-\left(n+1/2\right)\text{ln}n+n-M_n^{\left[m\right]}$$

Then

$$w_n-w_{n+1}=\left(n+1/2\right)\text{ln}\left(1+1/n\right)-1+M_{n+1}^{\left[m\right]}-M_n^{\left[m\right]}$$

By using the relations (19) and (22), we have

$$\begin{array}{ll}\hfill w_n-w_{n+1}& =\sum _{k=m+1}^{\infty }\frac{1}{2k+1}\frac{1}{{\left(2n+1\right)}^{2k}}-\frac{1}{4\left(2m+3\right)n\left(n+1\right){\left(2n+1\right)}^{2m}}\\ =\sum _{k=m+3}^{\infty }\frac{1}{2k+1}\frac{1}{{\left(2n+1\right)}^{2k}}-\frac{\left(5+2m+8n+8n^2\right){\left(2n+1\right)}^{-4-2m}}{4n\left(n+1\right)\left(15+16m+4m^2\right)}\end{array}$$

Then

$$\underset{n\to \infty }{\text{lim}}{n}^{2\left(m+2\right)}\left(w_n-w_{n+1}\right)=\frac{-1}{\left(2m+3\right)\left(2m+5\right)2^{2m+3}};n,m=1,2,3,....$$

(31)

Theorem 2. The rate of convergence of the sequence w _n is equal to n^-2m-3, since

$$\underset{n\to \infty }{\text{lim}}{n}^{2m+3}{w}_n=\frac{1}{{\left(2m+3\right)}^2\left(2m+5\right)2^{2m+3}}.$$

In 2011, Mortici [22] shows by numerical computations that his formula

$$n!~\sqrt{2\pi n}{\left(n/e\right)}^n{e}^{\frac{1}{12n+\frac{2}{5n}}}={\mu }_n$$

(32)

is much stronger than other known formulas such as:1

Table 1

The following table shows numerically that our new formula ${\lambda }_{n,1}=\sqrt{2\pi n}{\left(n/e\right)}^n{e}^{M_n^{\left[1\right]}}$ has a superiority over the the Mortici's formula μ _n .

Table 2