aa r X i v : . [ m a t h . F A ] J u l Inequalities for Euler-Mascheroni constant
Hongmin Xu and Xu YouJun 26, 2014
Abstract
The aim of this paper is to establish new inequalities for the Euler-Mascheroni bythe continued fraction method.
Euler’s Constant was first introduced by Leonhard Euler (1707-1783) in 1734 as the limitof the sequence γ ( n ) := n X m =1 m − ln n. (1.1)It is also known as the Euler-Mascheroni constant. There are many famous unsolvedproblems about the nature of this constant(See e.g. the survey papers or books of R.P.Brent and P. Zimmermann[1], Dence and Dence[3], Havil[5] and Lagarias[8]). For example,it is a long-standing open problem if it is a rational number. A good part of its mysterycomes from the fact that the known algorithms converging to γ are not very fast, at least,when they are compared to similar algorithms for π and e .The sequence ( r ( n )) n ∈ N converges very slowly toward γ , like (2 n ) − . Up to now, manyauthors are preoccupied to improve its rate of convergence, see e.g. [2, 3, 4, 6, 7, 9, 10,11, 14] and references therein. We list some main results as follows: n X m =1 m − ln (cid:18) n + 12 (cid:19) = γ + O ( n − ) , (DeTemple, [4]) n X m =1 m − ln n + n + + n + n + = γ + O ( n − ) , (Mortici, [11]) X m =1 m − ln (cid:18) n + 124 n − n + 235760 n (cid:19) = γ + O ( n − ) , (Chen and Mortici, [2])Recently, Mortici and Chen[14] provided a very interesting sequence ν ( n ) = n X m =1 m −
12 ln (cid:18) n + n + 13 (cid:19) − − (cid:0) n + n + (cid:1) + (cid:0) n + n + (cid:1) + + (cid:0) n + n + (cid:1) + (cid:0) n + n + (cid:1) ! , and proved lim n →∞ n ( ν ( n ) − γ ) = − . (1.2)Hence the rate of the convergence of the sequence ( ν ( n )) n ∈ N is n − .Very recently, by inserting the continued fraction term in (1.1), Lu[9] introduced aclass of sequences ( r k ( n )) n ∈ N (see Theorem 1 below), and showed172( n + 1) < γ − r ( n ) < n , (1.3) 1120( n + 1) < r ( n ) − γ < n − . (1.4)In fact, Lu[9] also found a without proof. In general, the continued fraction method couldprovide a better approximation than others, and has less numerical computations.First, we will prove Theorem 1.
For Euler-Mascheroni constant, we have the following convergent sequence r ( n ) = 1 + 12 + · · · + 1 n − ln n − a n + a nn + a nn + ... , where ( a , a , a , a , a , a , a ) = (cid:0) , , , , , , (cid:1) , and a k +1 = − a k for ≤ k ≤ .Let R k ( n ) := a n + a nn + a nn + a n ... n + ak , See Appendix for their simple expressions) and r k ( n ) := n X m =1 m − ln n − R k ( n ) . For ≤ k ≤ , we have lim n →∞ n k +1 ( r k ( n ) − γ ) = C k , (1.5) where ( C , · · · , C ) = (cid:0) − , − , , , − , − , , , − , − , , , − (cid:1) . Open problem
For every k ≥
1, we have a k +1 = − a k .The main aim of this paper is to improve (1.3) and (1.4). We establish the followingmore precise inequalities. Theorem 2.
Let r ( n ) , r ( n ) , C and C be defined in Theorem 1, then C n + 1) < γ − r ( n ) < C n , (1.6) C n + 1) < r ( n ) − γ < C n . (1.7) Remark . In fact, Theorem 2 implies that r ( n ) is a strictly increasing function of n , whereas r ( n ) is a strictly decreasing function of n . Certainly, it has the similarinequalities for r k ( n )(1 ≤ k ≤ r ( n ). Remark . It is worth to pointing out that Theorem 2 provides sharp bounds for harmonicsequence, which are superior to Theorem 3 and 4 of Mortici and Chen[14].
The following lemma gives a method for measuring the rate of convergence, This Lemmawas first used by Mortici[12, 13] for constructing asymptotic expansions, or to acceleratesome convergences. For proof and other details, see, e.g., [13].
Lemma 1.
If the sequence ( x n ) n ∈ N is convergent to zero and there exists the limit lim n → + ∞ n s ( x n − x n +1 ) = l ∈ [ −∞ , + ∞ ](2.1) with s > , then there exists the limit: lim n → + ∞ n s − x n = ls − . (2.2) 3n the sequel, we always assume n ≥ a ∈ R which produces the most accurate approximation ofthe form r ( n ) = n X m =1 m − ln n − a n , (2.3)here we note R ( n ) = a n . To measure the accuracy of this approximation, we usually saythat an approximation (2.3) is better as r ( n ) − γ faster converges to zero. Clearly r ( n ) − r ( n + 1) = ln (cid:18) n (cid:19) − n + 1 + a n + 1 − a n . (2.4)It is well-known that for | x | < x ) = ∞ X m =1 ( − m − x m m and 11 − x = ∞ X m =0 x m . Developing the expression (2.4) into power series expansion in n , we easily obtain r ( n ) − r ( n + 1) = (cid:18) − a (cid:19) n + (cid:18) a − (cid:19) n + (cid:18) − a (cid:19) n + O (cid:18) n (cid:19) . (2.5)From Lemma 1, we see that the rate of convergence of the sequence ( r ( n ) − γ ) n ∈ N iseven higher as the value s satisfying (2.1). By lemma 1, we have(i) If a = , then the rate of convergence of the ( r ( n ) − γ ) n ∈ N is n − , sincelim n →∞ n ( r ( n ) − γ ) = 12 − a = 0 . (ii) If a = , from (2.5) we have r ( n ) − r ( n + 1) = −
16 1 n + O (cid:18) n (cid:19) . Hence the rate of convergence of the ( r ( n ) − γ ) n ∈ N is n − , sincelim n →∞ n ( r ( n ) − γ ) = − . We also observe that the fastest possible sequence ( r ( n )) n ∈ N is obtained only for a = .Just as Lu[9] did, we may repeat the above approach to determine a to a step bystep. However, the computations become very difficult when k ≥
5. In this paper we willuse the
Mathematica software to manipulate symbolic computations.4et r k ( n ) = n X m =1 m − ln n − R k ( n ) , (2.6)then r k ( n ) − r k ( n + 1) = ln (cid:18) n (cid:19) − n + 1 + R k ( n + 1) − R k ( n ) . (2.7)It is easy to get the following power seriesln (cid:18) n (cid:19) − n + 1 = ∞ X m =2 ( − m m − m n m . (2.8)Hence the key step is to expand R k ( n + 1) − R k ( n ) into power series in n . Here we usesome examples to explain our method. Step 1: . For example, given a to a , find a . Define R ( n ) = n + n n + − n n + 35 ∗ nn + − ∗ nn + 79126 ∗ nn + − ∗ nn + a (2.9) = −
237 + 1405 a + 1800 n + 1740 a n − n + 3780 a n + 3780 n a + 600 a n + 600 n + 790 a n + 1260 a n + 1260 n ) . By using the
Mathematica software(The
Mathematica Program is very similar to onegiven in Remark 3 below, however it has a parameter a ), we obtain R ( n + 1) − R ( n )(2.10) = − n + 23 n − n + 45 n − n + 67 n − n + 360030 − a n + − a + 6241 a n + O (cid:18) n (cid:19) . Substituting (2.8) and (2.10) into (2.7), we get r ( n ) − r ( n + 1) = (cid:18) −
89 + 360030 − a (cid:19) n (2.11) + (cid:18)
910 + − a + 6241 a (cid:19) n + O (cid:18) n (cid:19) . r ( n )) n ∈ N is obtained only for a = . At the same time,it follows from (2.11), r ( n ) − r ( n + 1) = 580812446472 1 n + O (cid:18) n (cid:19) , (2.12)the rate of convergence of the ( r ( n ) − γ ) n ∈ N is n − , sincelim n →∞ n ( r ( n ) − γ ) = − . We can use the above approach to find a k (3 ≤ k ≤ a . Since a = − a , a = − a and a = − a . So we may conjecture a = − a .Now let’s check it carefully. Step 2:
Check a = − to a = − .Let a , · · · , a , and R ( n ) be defined in Theorem 1. Applying the Mathematica soft-ware, we obtain R ( n + 1) − R ( n )(2.13) = − n + 23 n − n + 45 n − n + 67 n − n + 89 1 n −
910 1 n + 736265836136 1 n + O (cid:18) n (cid:19) , which is the desired result. Substituting (2.8) and (2.13) into (2.7), we get r ( n ) − r ( n + 1) = − n + O (cid:18) n (cid:19) , (2.14)the rate of convergence of the ( r ( n ) − γ ) n ∈ N is n − , sincelim n →∞ n ( r ( n ) − γ ) = − . Next, we can use the
Step 1 to find a , and the Step a and a . Itshould be noted that Theorem 2 will provide their another proofs for a and a . So weomit the details here.Finally, we check a = − . R ( n + 1) − R ( n )(2.15) = − n + 23 n − n + 45 n − n + 67 n − n + 89 1 n −
910 1 n + 1011 1 n − n + 1213 1 n − n + 19036485866232576034146400 1 n + O (cid:18) n (cid:19) . r ( n ) − r ( n + 1) = − n + O (cid:18) n (cid:19) . (2.16)Since lim n →∞ n ( r ( n ) − γ ) = − , thus the rate of convergence of the ( r ( n ) − γ ) n ∈ N is n − .This completes the proof of Theorem 1. Remark . In fact, if the assertion a = − holds, then the other values a j (1 ≤ j ≤
12) must be true. The following
Mathematica program will generate R ( n +1) − R ( n )into power series in n with order 16:Normal[Series[( R [ n + 1] − R [ n ]) /. n → /x, { x, , } ]] /. x → /n Remark . It is a very interesting question to find a k for k ≥
14. However, it seemsimpossible by the above method.
Before we prove the Theorem 2, let us give a simple inequality, which plays an importantrole of the proof.
Lemma 2.
Let f ′′ ( x ) be a continuous function. If f ′′ ( x ) > , then Z a +1 a f ( x ) dx > f ( a + 1 / . (3.1) Proof.
Let x = a + 1 /
2. By Taylor’s formula, we have Z a +1 a f ( x ) dx = Z a +1 a (cid:18) f ( x ) + f ′ ( x )( x − x ) + 12 f ′′ ( θ x )( x − x ) (cid:19) dx> Z a +1 a (cid:0) f ( x ) + f ′ ( x )( x − x ) (cid:1) dx = f ( a + 1 / . This completes the proof of Lemma 2.In the sequel, the notation P k ( x ) means a polynomial of degree k in x with all of itsnon-zero coefficients positive, which may be different at each occurrence.Let’s begin to prove Theorem 2. Note r ( ∞ ) = 0, it is easy to see γ − r ( n ) = ∞ X m = n ( r ( m + 1) − r ( m )) = ∞ X m = n f ( m ) , (3.2) 7here f ( m ) = 1 m + 1 − ln (cid:18) m (cid:19) − R ( m + 1) + R ( m ) . Let D = . By using the Mathematica software, we have f ′ ( x ) + D x + 1) = − P ( x )( x −
1) + 1619906998377 · · · x (1 + x ) P (1)10 ( x ) P (2)10 ( x ) < , and f ′ ( x ) + D (cid:0) x + (cid:1) = P ( x )4226263965 x (1 + x ) (1 + 2 x ) P (3)10 ( x ) P (4)10 ( x ) > . Hence, we get the following inequalities for x ≥ D x + 1) < − f ′ ( x ) < D x + ) (3.3)Applying f ( ∞ ) = 0, (3.3) and Lemma 2, we get f ( m ) = − Z ∞ m f ′ ( x ) dx ≤ D Z ∞ m (cid:18) x + 12 (cid:19) − dx (3.4) = D (cid:18) m + 12 (cid:19) − ≤ D Z m +1 m x − dx. From (3.1) and (3.4) we obtain γ − r ( n ) ≤ ∞ X m = n D Z m +1 m x − dx (3.5) = D Z ∞ n x − dx = D
132 1 n . Similarly, we also have f ( m ) = − Z ∞ m f ′ ( x ) dx ≥ D Z ∞ m ( x + 1) − dx = D
12 ( m + 1) − ≥ D Z m +2 m +1 x − dx, and γ − r ( n ) ≥ ∞ X m = n D Z m +2 m +1 x − dx (3.6) = D Z ∞ n +1 x − dx = D
132 1( n + 1) . r ( ∞ ) = 0, it is easy to deduce r ( n ) − γ = ∞ X m = n ( r ( m ) − r ( m + 1)) = ∞ X m = n g ( m ) , (3.7)where g ( m ) = ln (cid:18) m (cid:19) − m + 1 − R ( m ) + R ( m + 1) . We write D = . By using the Mathematica software, we have − g ′ ( x ) − D x + 1) = P ( x )24495240 x (1 + x ) P (1)8 ( x ) P (2)8 ( x ) > − g ′ ( x ) − D (cid:0) x + (cid:1) = − P ( x )( x −
1) + 46220056778393539977246763077416123810 x (1 + x ) (1 + 2 x ) P (3)8 ( x ) P (4)8 ( x ) < . Hence for x ≥ D x + 1) < − g ′ ( x ) < D (cid:0) x + (cid:1) . (3.8)Applying g ( ∞ ) = 0, (3.8) and (3.1), we get g ( m ) = − Z ∞ m g ′ ( x ) dx ≤ D Z ∞ m (cid:18) x + 12 (cid:19) − dx (3.9) = D (cid:18) m + 12 (cid:19) − ≤ D Z m +1 m x − dx. It follows from (3.7) and (3.9) r ( n ) − γ ≤ ∞ X m = n D Z m +1 m x − dx (3.10) = D Z ∞ n x − dx = D
156 1 n . Finally, g ( m ) = − Z ∞ m g ′ ( x ) dx ≥ D Z ∞ m ( x + 1) − dx = D
13 ( m + 1) − ≥ D Z m +2 m +1 x − dx. r ( n ) − γ ≥ ∞ X m = n D Z m +2 m +1 x − dx (3.11) = D Z ∞ n +1 x − dx = D
156 1( n + 1) . Combining (3.10) and (3.11) completes the proof of (1.7).
Remark . As an example, we give the
Mathematica Program for the proof of the left-handside of (3.3):(i) Together[D[ f [ x ] , { x, } ]+ D ( x + 1) ];(ii) Take out the numerator P [ x ] of the above rational function, then manipulate theprogram: Apart[ P [ x ] / ( x − The authors declare that they have no competing interests.
Hongmin Xu conceived of the algorithm and helped to draft the manuscript. Xu Youcarried out the design of the program and drafted the manuscript. All authors read andapproved the final manuscript.
The authors thank Prof. Xiaodong Cao for his help.This research of this paper was supported by the National Natural Science Foundationof China (Grant No.11171344) and the Natural Science Foundation of Beijing (GrantNo.1112010).
Appendix
For the reader’s convenience, we rewrite R k ( n ) ( k ≤
13) with minimal10enominators as following. R ( n ) = 12 n ,R ( n ) = 12 n −