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Research | Open | Published:

Continued fraction inequalities for the Euler-Mascheroni constant

Abstract

The aim of this paper is to establish new inequalities for the Euler-Mascheroni constant by the continued fraction method.

MSC:11Y60, 41A25, 41A20.

1 Introduction

The Euler-Mascheroni constant was first introduced by Leonhard Euler (1707-1783) in 1734 as the limit of the sequence

γ(n):= m = 1 n 1 m lnn.
(1.1)

There are many famous unsolved problems about the nature of this constant (see, e.g., the survey papers or books of Brent and Zimmermann [1], Dence and Dence [2], Havil [3] and Lagarias [4]). For example, it is a long-standing open problem if the Euler-Mascheroni constant is a rational number. A good part of its mystery comes 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 ( γ ( n ) ) n N converges very slowly toward γ, like ( 2 n ) 1 . Up to now, many authors have been preoccupied with improving its rate of convergence (see, e.g., [2, 522] and the references therein). We list some main results as follows:

m = 1 n 1 m ln ( n + 1 2 ) = γ + O ( n 2 ) (DeTemple [6]) , m = 1 n 1 m ln n 3 + 3 2 n 2 + 227 240 + 107 480 n 2 + n + 97 240 = γ + O ( n 6 ) (Mortici [13]) , m = 1 n 1 m ln ( 1 + 1 2 n + 1 24 n 2 1 48 n 3 + 23 5 , 760 n 4 ) = γ + O ( n 5 ) (Chen and Mortici [5]).

Recently, Mortici and Chen [14] provided a very interesting sequence,

ν ( n ) = m = 1 n 1 m 1 2 ln ( n 2 + n + 1 3 ) ( 1 180 ( n 2 + n + 1 3 ) 2 + 8 2 , 835 ( n 2 + n + 1 3 ) 3 + + 5 1 , 512 ( n 2 + n + 1 3 ) 4 + 592 93 , 555 ( n 2 + n + 1 3 ) 5 ) ,

and proved

lim n n 12 ( ν ( n ) γ ) = 796 , 801 43 , 783 , 740 .
(1.2)

Hence the rate of convergence of the sequence ( ν ( n ) ) n N is n 12 .

Very recently, by inserting the continued fraction term in (1.1), Lu [9] introduced a class of sequences ( r k ( n ) ) n N (see Theorem 1) and showed

1 72 ( n + 1 ) 3 <γ r 2 (n)< 1 72 n 3 ,
(1.3)
1 120 ( n + 1 ) 4 < r 3 (n)γ< 1 120 ( n 1 ) 4 .
(1.4)

In fact, Lu [9] also found a 4 without proof. In general, the continued fraction method could provide a better approximation than others, and has less numerical computations.

First, we will prove the following theorem.

Theorem 1 For the Euler-Mascheroni constant, we have the following convergent sequence:

r(n)=1+ 1 2 ++ 1 n lnn a 1 n + a 2 n n + a 3 n n + ,

where ( a 1 , a 2 , a 4 , a 6 , a 8 , a 10 , a 12 )=( 1 2 , 1 6 , 3 5 , 79 126 , 7 , 230 6 , 241 , 4 , 146 , 631 3 , 833 , 346 , 306 , 232 , 774 , 533 179 , 081 , 182 , 865 ), and a 2 k + 1 = a 2 k for 1k6.

Let

R k (n):= a 1 n + a 2 n n + a 3 n n + a 4 n n + a k

(see the Appendix for their simple expressions) and

r k (n):= m = 1 n 1 m lnn R k (n).

For 1k13, we have

lim n n k + 1 ( r k ( n ) γ ) = C k ,
(1.5)

where

( C 1 , , C 13 ) = ( 1 12 , 1 72 , 1 120 , 1 200 , 79 25 , 200 , 6 , 241 3 , 175 , 200 , 241 105 , 840 , 58 , 081 22 , 018 , 248 , 262 , 445 91 , 974 , 960 , 2 , 755 , 095 , 121 892 , 586 , 949 , 408 , 20 , 169 , 451 3 , 821 , 257 , 440 , 406 , 806 , 753 , 641 , 401 45 , 071 , 152 , 103 , 463 , 200 , 71 , 521 , 421 , 431 5 , 152 , 068 , 292 , 800 ) .

Open problem For every k1, we have a 2 k + 1 = a 2 k .

The main aim of this paper is to improve (1.3) and (1.4). We establish the following more precise inequalities.

Theorem 2 Let r 10 (n), r 11 (n), C 10 and C 11 be defined in Theorem  1, then

C 10 1 ( n + 1 ) 11 <γ r 10 (n)< C 10 1 n 11 ,
(1.6)
C 11 1 ( n + 1 ) 12 < r 11 (n)γ< C 11 1 n 12 .
(1.7)

Remark 1 In fact, Theorem 2 implies that r 10 (n) is a strictly increasing function of n, whereas r 11 (n) is a strictly decreasing function of n. Certainly, it has similar inequalities for r k (n) (1k9), we leave these for readers to verify. It is also should be noted that (1.4) cannot deduce the monotonicity of r 3 (n).

Remark 2 It is worth to point out that Theorem 2 provides sharp bounds for a harmonic sequence which are superior to Theorems 3 and 4 of Mortici and Chen [14].

2 The proof of Theorem 1

The following lemma gives a method for measuring the rate of convergence. This lemma was first used by Mortici [23, 24] for constructing asymptotic expansions or to accelerate some convergences. For proof and other details, see, e.g., [24].

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>1, then there exists the limit

lim n + n s 1 x n = l s 1 .
(2.2)

In the sequel, we always assume n2.

We need to find the value a 1 R which produces the most accurate approximation of the form

r 1 (n)= m = 1 n 1 m lnn a 1 n ,
(2.3)

here we note R 1 (n)= a 1 /n. To measure the accuracy of this approximation, we usually say that approximation (2.3) is better as r 1 (n)γ faster converges to zero. Clearly,

r 1 (n) r 1 (n+1)=ln ( 1 + 1 n ) 1 n + 1 + a 1 n + 1 a 1 n .
(2.4)

It is well known that for |x|<1,

ln(1+x)= m = 1 ( 1 ) m 1 x m m and 1 1 x = m = 0 x m .

Developing expression (2.4) into power series expansion in 1/n, we obtain

r 1 (n) r 1 (n+1)= ( 1 2 a 1 ) 1 n 2 + ( a 1 2 3 ) 1 n 3 + ( 3 4 a 1 ) 1 n 4 +O ( 1 n 5 ) .
(2.5)

From Lemma 1, we see that the rate of convergence of the sequence ( r 1 ( n ) γ ) n N is even higher than the value s satisfying (2.1). By Lemma 1, we have

  1. (i)

    If a 1 1 2 , then the rate of convergence of ( r 1 ( n ) γ ) n N is n 1 since

    lim n n ( r 1 ( n ) γ ) = 1 2 a 1 0.
  2. (ii)

    If a 1 = 1 2 , from (2.5) we have

    r 1 (n) r 1 (n+1)= 1 6 1 n 3 +O ( 1 n 4 ) .

Hence the rate of convergence of ( r 1 ( n ) γ ) n N is n 2 since

lim n n 2 ( r 1 ( n ) γ ) = 1 12 .

We also observe that the fastest possible sequence ( r 1 ( n ) ) n N is obtained only for a 1 = 1 2 .

Just as Lu [9] did, we may repeat the above approach to determine a 1 to a 4 step by step. However, the computations become very difficult when k5. In this paper we use Mathematica software to manipulate symbolic computations.

Let

r k (n)= m = 1 n 1 m lnn R k (n),
(2.6)

then

r k (n) r k (n+1)=ln ( 1 + 1 n ) 1 n + 1 + R k (n+1) R k (n).
(2.7)

It is easy to get the following power series:

ln ( 1 + 1 n ) 1 n + 1 = m = 2 ( 1 ) m m 1 m 1 n m .
(2.8)

Hence the key step is to expand R k (n+1) R k (n) into power series in 1 n . Here we use some examples to explain our method.

Step 1: For example, given a 1 to a 7 , find a 8 . Define

R 8 ( n ) = 1 2 n + n 6 n + n 6 n + 3 5 n n + 3 5 n n + 79 126 n n + 79 126 n n + a 8 = 237 + 1 , 405 a 8 + 1 , 800 n + 1 , 740 a 8 n 630 n 2 + 3 , 780 a 8 n 2 + 3 , 780 n 3 6 ( 79 a 8 + 600 a 8 n + 600 n 2 + 790 a 8 n 2 + 1 , 260 a 8 n 3 + 1 , 260 n 4 ) .
(2.9)

By using Mathematica software (Mathematica Program is very similar to the one given in Remark 3; however, it has a parameter a 8 ), we obtain

R 8 ( n + 1 ) R 8 ( n ) = 1 2 n 2 + 2 3 n 3 3 4 n 4 + 4 5 n 5 5 6 n 6 + 6 7 n 7 7 8 n 8 + 360 , 030 6 , 241 a 8 396 , 900 n 9 + 346 , 440 + 24 , 964 a 8 + 6 , 241 a 8 2 352 , 800 n 10 + O ( 1 n 11 ) .
(2.10)

Substituting (2.8) and (2.10) into (2.7), we get

r 8 ( n ) r 8 ( n + 1 ) = ( 8 9 + 360 , 030 6 , 241 a 8 396 , 900 ) 1 n 9 + ( 9 10 + 346 , 440 + 24 , 964 a 8 + 6 , 241 a 8 2 352 , 800 ) 1 n 10 + O ( 1 n 11 ) .
(2.11)

The fastest possible sequence ( r 8 ( n ) ) n N is obtained only for a 8 = 7 , 230 6 , 241 . At the same time, it follows from (2.11) that

r 8 (n) r 8 (n+1)= 58 , 081 2 , 446 , 472 1 n 10 +O ( 1 n 11 ) ,
(2.12)

the rate of convergence of ( r 8 ( n ) γ ) n N is n 9 since

lim n n 9 ( r 8 ( n ) γ ) = 58 , 081 22 , 018 , 248 .

We can use the above approach to find a k (3k8). Unfortunately, it does not work well for a 9 . Since a 3 = a 2 , a 5 = a 4 and a 7 = a 6 . So, we may conjecture a 9 = a 8 . Now let us check it carefully.

Step 2: Check a 9 = 7 , 230 6 , 241 to a 13 = 306 , 232 , 774 , 533 179 , 081 , 182 , 865 .

Let a 1 ,, a 9 and R 9 (n) be defined in Theorem 1. Applying Mathematica software, we obtain

R 9 ( n + 1 ) R 9 ( n ) = 1 2 n 2 + 2 3 n 3 3 4 n 4 + 4 5 n 5 5 6 n 6 + 6 7 n 7 7 8 n 8 + 8 9 1 n 9 9 10 1 n 10 + 736 , 265 836 , 136 1 n 11 + O ( 1 n 12 ) ,
(2.13)

which is the desired result. Substituting (2.8) and (2.13) into (2.7), we get

r 9 (n) r 9 (n+1)= 262 , 445 9 , 197 , 496 1 n 11 +O ( 1 n 12 ) ,
(2.14)

the rate of convergence of ( r 9 ( n ) γ ) n N is n 10 since

lim n n 10 ( r 9 ( n ) γ ) = 262 , 445 91 , 974 , 960 .

Next, we can use Step 1 to find a 10 , and Step 2 to check a 11 and a 12 . It should be noted that Theorem 2 will provide the other proofs for a 10 and a 11 . So we omit the details here.

Finally, we check a 13 = 306 , 232 , 774 , 533 179 , 081 , 182 , 865 .

R 13 ( n + 1 ) R 13 ( n ) = 1 2 n 2 + 2 3 n 3 3 4 n 4 + 4 5 n 5 5 6 n 6 + 6 7 n 7 7 8 n 8 + 8 9 1 n 9 9 10 1 n 10 + 10 11 1 n 11 11 12 1 n 12 + 12 13 1 n 13 13 14 1 n 14 + 1 , 903 , 648 , 586 , 623 2 , 576 , 034 , 146 , 400 1 n 15 + O ( 1 n 16 ) .
(2.15)

Substituting (2.8) and (2.15) into (2.7), one has

r 13 (n) r 13 (n+1)= 500 , 649 , 950 , 017 2 , 576 , 034 , 146 , 400 1 n 15 +O ( 1 n 16 ) .
(2.16)

Since

lim n n 14 ( r 13 ( n ) γ ) = 71 , 521 , 421 , 431 5 , 152 , 068 , 292 , 800 ,

thus the rate of convergence of ( r 13 ( n ) γ ) n N is n 14 .

This completes the proof of Theorem 1.

Remark 3 In fact, if the assertion a 13 = 306 , 232 , 774 , 533 179 , 081 , 182 , 865 holds, then the other values a j (1j12) must be true. The following Mathematica Program will generate R 13 (n+1) R 13 (n) into power series in 1 n with order 16: Normal[Series[( R 13 [n+1] R 13 [n])/.n1/x,{x,0,16}]]/. x1/n.

Remark 4 It is a very interesting question to find a k for k14. However, it seems impossible by the above method.

3 The proof of Theorem 2

Before we prove Theorem 2, let us give a simple inequality by the Hermite-Hadamard inequality, which plays an important role in the proof.

Lemma 2 Let f be twice derivable with f continuous. If f (x)>0, then

a a + 1 f(x)dx>f(a+1/2).
(3.1)

In the sequel, the notation P k (x) means a polynomial of degree k in x with all of its non-zero coefficients positive, which may be different at each occurrence.

Let us begin to prove Theorem 2. Note r 10 ()=0, it is easy to see

γ r 10 (n)= m = n ( r 10 ( m + 1 ) r 10 ( m ) ) = m = n f(m),
(3.2)

where

f(m)= 1 m + 1 ln ( 1 + 1 m ) R 10 (m+1)+ R 10 (m).

Let D 1 = 2 , 755 , 095 , 121 6 , 762 , 022 , 344 . By using Mathematica software, we have

f (x)+ D 1 1 ( x + 1 ) 13 = P 19 ( x ) ( x 1 ) + 1 , 619 , 906 , 998 , 377 5 , 270 , 931 33 , 810 , 111 , 720 x ( 1 + x ) 13 P 10 ( 1 ) ( x ) P 10 ( 2 ) ( x ) <0,

and

f (x)+ D 1 1 ( x + 1 2 ) 13 = P 22 ( x ) 4 , 226 , 263 , 965 x ( 1 + x ) 2 ( 1 + 2 x ) 13 P 10 ( 3 ) ( x ) P 10 ( 4 ) ( x ) >0.

Hence, we get the following inequalities for x1:

D 1 1 ( x + 1 ) 13 < f (x)< D 1 1 ( x + 1 2 ) 13 .
(3.3)

Applying f()=0, (3.3) and Lemma 2, we get

f ( m ) = m f ( x ) d x D 1 m ( x + 1 2 ) 13 d x = D 1 12 ( m + 1 2 ) 12 D 1 12 m m + 1 x 12 d x .
(3.4)

From (3.1) and (3.4) we obtain

γ r 10 ( n ) m = n D 1 12 m m + 1 x 12 d x = D 1 12 n x 12 d x = D 1 132 1 n 11 .
(3.5)

Similarly, we also have

f ( m ) = m f ( x ) d x D 1 m ( x + 1 ) 13 d x = D 1 12 ( m + 1 ) 12 D 1 12 m + 1 m + 2 x 12 d x

and

γ r 10 ( n ) m = n D 1 12 m + 1 m + 2 x 12 d x = D 1 12 n + 1 x 12 d x = D 1 132 1 ( n + 1 ) 11 .
(3.6)

Combining (3.5) and (3.6) completes the proof of (1.6).

Note r 11 ()=0, it is easy to deduce

r 11 (n)γ= m = n ( r 11 ( m ) r 11 ( m + 1 ) ) = m = n g(m),
(3.7)

where

g(m)=ln ( 1 + 1 m ) 1 m + 1 R 11 (m)+ R 11 (m+1).

We write D 2 = 20 , 169 , 451 24 , 495 , 240 . By using Mathematica software, we have

g (x) D 2 1 ( x + 1 ) 14 = P 18 ( x ) 24 , 495 , 240 x 3 ( 1 + x ) 14 P 8 ( 1 ) ( x ) P 8 ( 2 ) ( x ) >0

and

g ( x ) D 2 1 ( x + 1 2 ) 14 = P 19 ( x ) ( x 1 ) + 4 , 622 , 005 , 677 , 839 , 353 , 997 , 724 , 676 , 307 , 741 6 , 123 , 810 x 3 ( 1 + x ) 3 ( 1 + 2 x ) 14 P 8 ( 3 ) ( x ) P 8 ( 4 ) ( x ) < 0 .

Hence, for x1,

D 2 1 ( x + 1 ) 14 < g (x)< D 2 1 ( x + 1 2 ) 14 .
(3.8)

Applying g()=0, (3.8) and (3.1), we get

g ( m ) = m g ( x ) d x D 2 m ( x + 1 2 ) 14 d x = D 2 13 ( m + 1 2 ) 13 D 2 13 m m + 1 x 13 d x .
(3.9)

It follows from (3.7) and (3.9) that

r 11 ( n ) γ m = n D 2 13 m m + 1 x 13 d x = D 2 13 n x 13 d x = D 2 156 1 n 12 .
(3.10)

Finally,

g ( m ) = m g ( x ) d x D 2 m ( x + 1 ) 14 d x = D 2 13 ( m + 1 ) 13 D 2 13 m + 1 m + 2 x 13 d x

and

r 11 ( n ) γ m = n D 2 13 m + 1 m + 2 x 13 d x = D 2 13 n + 1 x 13 d x = D 2 156 1 ( n + 1 ) 12 .
(3.11)

Combining (3.10) and (3.11) completes the proof of (1.7).

Remark 5 As an example, we give Mathematica Program for the proof of the left-hand side of (3.3):

  1. (i)

    Together [D[f[x],{x,1}]+ D 1 ( x + 1 ) 13 ];

  2. (ii)

    Take out the numerator P[x] of the above rational function, then manipulate the program: Apart [P[x]/(x1)].

Appendix

For the reader’s convenience, we rewrite R k (n) (k13) with minimal denominators as follows.

R 1 ( n ) = 1 2 n , R 3 ( n ) = 1 2 n 1 12 1 n 2 , R 5 ( n ) = 1 2 n 5 6 ( 1 + 10 n 2 ) , R 7 ( n ) = 1 2 n 79 1 , 200 1 n 2 147 400 ( 10 + 21 n 2 ) , R 9 ( n ) = 1 2 n 7 ( 871 + 790 n 2 ) 20 ( 241 + 3 , 990 n 2 + 3 , 318 n 4 ) , R 11 ( n ) = 1 2 n 52 , 489 894 , 348 1 n 2 1 , 237 , 227 , 621 + 584 , 280 , 400 n 2 4 , 471 , 740 ( 3 , 549 + 13 , 020 n 2 + 5 , 302 n 4 ) , R 13 ( n ) = 1 2 n 39 , 577 , 260 , 671 + 66 , 288 , 226 , 620 n 2 + 15 , 762 , 446 , 700 n 4 1 , 260 ( 20 , 169 , 451 + 434 , 410 , 620 n 2 + 646 , 328 , 298 n 4 + 150 , 118 , 540 n 6 ) , R 2 ( n ) = 3 6 n + 1 , R 4 ( n ) = 13 + 30 n 6 ( 1 + 6 n + 10 n 2 ) , R 6 ( n ) = 5 ( 281 + 348 n + 756 n 2 ) 6 ( 79 + 600 n + 790 n 2 + 1 , 260 n 3 ) , R 8 ( n ) = 964 , 337 + 2 , 646 , 000 n + 2 , 599 , 730 n 2 + 2 , 621 , 220 n 3 20 ( 19 , 039 + 144 , 600 n + 315 , 210 n 2 + 303 , 660 n 3 + 262 , 122 n 4 ) , R 10 ( n ) = ( 7 ( 108 , 237 , 701 + 208 , 886 , 046 n + 523 , 341 , 290 n 2 R 10 ( n ) = + 210 , 464 , 400 n 3 + 230 , 000 , 760 n 4 ) ) R 10 ( n ) = / ( 20 ( 12 , 649 , 849 + 107 , 768 , 934 n + 209 , 431 , 110 n 2 R 10 ( n ) = + 395 , 365 , 320 n 3 + 174 , 158 , 502 n 4 + 161 , 000 , 532 n 5 ) ) , R 12 ( n ) = ( 3 , 604 , 759 , 235 , 968 , 501 + 11 , 032 , 319 , 618 , 513 , 046 n R 12 ( n ) = + 17 , 366 , 281 , 558 , 290 , 420 n 2 + 19 , 958 , 033 , 982 , 902 , 400 n 3 R 12 ( n ) = + 7 , 661 , 417 , 445 , 218 , 460 n 4 + 4 , 964 , 130 , 389 , 017 , 800 n 5 ) R 12 ( n ) = / ( 1 , 260 ( 1 , 058 , 674 , 313 , 539 + 9 , 019 , 254 , 081 , 474 n R 12 ( n ) = + 22 , 801 , 779 , 033 , 180 n 2 + 33 , 088 , 387 , 754 , 520 n 3 + 33 , 925 , 126 , 033 , 722 n 4 R 12 ( n ) = + 13 , 474 , 242 , 079 , 452 n 5 + 7 , 879 , 572 , 046 , 060 n 6 ) ) .

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Acknowledgements

The authors would like to thank the referees and Prof Xiaodong Cao for their careful reading of the manuscript and insightful comments. Research of this paper was supported by the National Natural Science Foundation of China (Grant No.11171344) and the Natural Science Foundation of Beijing (Grant No.1112010).

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Correspondence to Xu You.

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Keywords

  • Euler-Mascheroni constant
  • rate of convergence
  • continued fraction
  • Taylor’s formula
  • harmonic sequence