SOME INEQUALITIES REGARDING A GENERALIZATION OF EULER’S CONSTANT

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Generalization of Euler’s Constant Alina Sînt ˘am ˘arian vol. 9, iss. 2, art. 46, 2008

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SOME INEQUALITIES REGARDING A GENERALIZATION OF EULER’S CONSTANT

ALINA SÎ NT ˘AM ˘ARIAN

Department of Mathematics Technical University of Cluj-Napoca Str. C. Daicoviciu nr. 15

400020 Cluj-Napoca, Romania.

EMail:Alina.Sintamarian@math.utcluj.ro

Received: 28 November, 2007

Accepted: 20 March, 2008

Communicated by: L. Tóth 2000 AMS Sub. Class.: 11Y60, 40A05.

Key words: Sequence, Convergence, Euler’s constant, Approximation, Estimate, Series.

Abstract: The purpose of this paper is to evaluate the limitγ(a)of the sequence

1

a + 1

a+ 1+· · ·+ 1

a+n−1−lna+n−1 a

n∈N

,

wherea∈(0,+∞). We give some lower and upper estimates for 1

a+ 1

a+ 1+· · ·+ 1

a+n−1−lna+n−1

a −γ(a), n∈N.

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1. Introduction

Let (D_n)n∈N be the sequence defined by D_n = 1 + ¹₂ +· · ·+ _n¹ −lnn, for each n ∈ N. It is well-known that the sequence (D_n)n∈N is convergent and its limit, usually denoted byγ, is called Euler’s constant.

ForDn−γ, n ∈ N, many lower and upper estimates have been obtained in the literature. We recall some of them:

• _2(n+1)¹ < D_n−γ < _2(n−1)¹ , for eachn∈N\ {1}([14]);

• _2(n+1)¹ < D_n−γ < _2n¹ , for eachn∈N([8], [19]);

• _2n+1¹ < D_n−γ < _2n¹ , for eachn∈N([17]);

• _2n+¹ 2 5

< D_n−γ < _2n+¹ 1 3

, for eachn ∈N([15], [16]);

• ¹

2n+^2γ−1_1−γ ≤D_n−γ < ¹

2n+¹₃, for eachn∈N([16, Editorial comment], [2], [3]).

In Section 2 we present a generalization of Euler’s constant as the limit of the sequence

1 a + 1

a+ 1 +· · ·+ 1

n∈N

, a∈(0,+∞), and we denote this limit byγ(a).

In Section3we give some lower and upper estimates for 1

a + 1

a+ 1 +· · ·+ 1

a+n−1 −lna+n−1

a −γ(a), n∈N.

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2. The Number γ(a)

It is known that the sequence 1

a + 1

a+ 1 +· · ·+ 1

n∈N

, a∈(0,+∞), is convergent (see for example [5, p. 453], [7], where problems in this sense were proposed; [6]; [13]).

The results contained in the following theorem were given in [10].

Theorem 2.1. Leta ∈ (0,+∞). We consider the sequences (x_n)n∈N and(y_n)n∈N

defined by

x_n = 1 a + 1

a+ 1 +· · ·+ 1

a+n−1 −lna+n a and

y_n= 1 a + 1

a+ 1 +· · ·+ 1

a+n−1−lna+n−1

a ,

for eachn ∈N. Then:

(i) the sequences(x_n)n∈Nand(y_n)n∈Nare convergent to the same number, which we denote byγ(a), and satisfy the inequalitiesx_n < x_n+1 < γ(a)< y_n+1 < y_n, for eachn∈N;

(ii) 0< _a¹ −ln 1 + ¹_a

< γ(a)< ¹_a; (iii) lim

n→∞n(γ(a)−x_n) = ¹₂ and lim

n→∞n(y_n−γ(a)) = ¹₂.

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Remark 1. The sequence (y_n)_n∈_N from Theorem 2.1, for a = 1, becomes the sequence(D_n)n∈N, soγ(1) =γ.

The following theorem was given by the author in [12, Theorem 2.3].

Theorem 2.2. Let a ∈ (0,+∞). We consider the sequence (u_n)n∈N defined by u_n = y_n− _2(a+n−1)+¹ 1

3

, for each n ∈ N, where (y_n)n∈N is the sequence from the statement of Theorem 2.1. Also, we specify that γ(a) is the limit of the sequence (y_n)n∈N.

Then:

(i) un < un+1 < γ(a), for eachn∈N\ {1}, and lim

n→∞n³(γ(a)−un) = ₇₂¹; (ii) _2(a+n−1)+¹ 11

28

< y_n−γ(a)< _2(a+n−1)+¹ 1 3

, for eachn∈N\ {1}.

Remark 2. The lower estimate from part(ii)of Theorem2.2holds forn= 1as well.

Remark 3. The second limit from part(iii)of Theorem 2.1also follows from part (ii)of Theorem2.2.

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3. Proving Some Estimates for y

_n

− γ(a) using the Logarithmic Derivative of the Gamma Function

As we already mentioned in Section1, it is known that ([16, Editorial comment], [2, Theorem 3], [3, Theorem 1.1])

1

2n+ ^2γ−1_1−γ ≤D_n−γ < 1 2n+¹₃,

for eachn∈N, the constants ^2γ−1_1−γ and ¹₃ being the best possible with this property.

Let a ∈ (0,+∞). In a similar way as in the proof given by H. Alzer in [2, Theorem 3], we shall obtain lower and upper estimates fory_n−γ(a)(n∈N), where (y_n)n∈N is the sequence from the statement of Theorem2.1, the limit of which we denoted byγ(a). In order to do this we shall prove, in a similar way as in [3, Lemma 2.1], some finer inequalities than those used by H. Alzer in [2, Theorem 3].

Lemma 3.1. We have:

(i) ψ(x+ 1)−lnx > _2x¹ − _12x¹2 +_120x¹ 4 −_252x¹ 6, for eachx∈(0,+∞);

(ii) ¹_x −ψ⁰(x+ 1)< _2x¹2 − _6x¹3 + _30x¹5 −_42x¹7 + _30x¹9, for eachx∈(0,+∞).

We specify that the functionψ is the logarithmic derivative of the gamma function, i.e.ψ(x) = ^Γ_Γ(x)⁰^(x), for eachx∈(0,+∞).

Proof. (i)It is known (see, for example, [18, p. 116]) thatlnx=R∞ 0

e^−t−e^−xt t dt, for eachx∈(0,+∞). Also, we shall need the formula

ψ(x) = Z ∞

0

e^−t

t − e^−xt 1−e^−t

dt,

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which holds for each x ∈ (0,+∞), known as Gauss’ expression of ψ(x) as an infinite integral (see, for example, [18, p. 247]). Having in view the above relations, we are able to write that

ψ(x+ 1)−lnx= Z ∞

0

1

t − 1 e^t−1

e^−xtdt,

for eachx∈(0,+∞).

It is not difficult to verify that Z ∞

0

tⁿe^−xtdt = n!

xⁿ⁺¹, for eachn∈N∪ {0}, anyx∈(0,+∞).

Then we have ψ(x+ 1)−lnx− 1

2x + 1

12x² − 1

120x⁴ + 1 252x⁶

= Z ∞

0

1

t − 1

e^t−1 − 1 2+ t

12− t³

720 + t⁵ 30240

e^−xtdt

= Z ∞

0

1

30240t(e^t−1)[30240(e^t−1)−30240t−15120t(e^t−1) + 2520t²(e^t−1)

−42t⁴(e^t−1) +t⁶(e^t−1)]e^−xtdt

= Z ∞

0

1 30240t(e^t−1)

"

30240

∞

X

n=2

tⁿ

n!−15120

∞

X

n=1

tⁿ⁺¹

n! + 2520

∞

X

n=1

tⁿ⁺² n!

−42

∞

X

n=1

tⁿ⁺⁴ n! +

∞

X

n=1

tⁿ⁺⁶ n!

#

e^−xtdt

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= Z ∞

0

∞

P

n=9

(n−3)(n−5)(n−7)(n−8)(n²+8n+36)

n! tⁿ

30240t(e^t−1) ·e^−xtdt >0, for eachx∈(0,+∞).

(ii)In part(i)we obtained that lnx−ψ(x+ 1) =

Z ∞

0

1

e^t−1− 1 t

e^−xtdt,

for eachx∈(0,+∞). Differentiating here we get that 1

x−ψ⁰(x+ 1) = Z ∞

0

1− t e^t−1

e^−xtdt,

for eachx∈(0,+∞).

Then we have 1

x−ψ⁰(x+ 1)− 1

2x² + 1

6x³ − 1

30x⁵ + 1

42x⁷ − 1 30x⁹

= Z ∞

0

1− t

e^t−1− t 2 + t²

12− t⁴

720 + t⁶

30240− t⁸ 1209600

e^−xtdt

= Z ∞

0

1

1209600(e^t−1)[1209600(e^t−1)−1209600t−604800t(e^t−1)

+ 100800t²(e^t−1)−1680t⁴(e^t−1) + 40t⁶(e^t−1)−t⁸(e^t−1)]e^−xtdt

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= Z ∞

0

1

1209600(e^t−1)

"

1209600

∞

X

n=2

tⁿ

n!−604800

∞

X

n=1

tⁿ⁺¹ n!

+100800

∞

X

n=1

tⁿ⁺²

n! −1680

∞

X

n=1

tⁿ⁺⁴ n! + 40

∞

X

n=1

tⁿ⁺⁶ n! −

∞

X

n=1

tⁿ⁺⁸ n!

#

e^−xtdt

=− Z ∞

0

P∞ n=11

(n−3)(n−5)(n−7)(n−9)(n−10)(n+4)(n²+2n+32)

n! tⁿ

1209600(e^t−1) ·e^−xtdt <0, for eachx∈(0,+∞).

Remark 4. In fact, these inequalities from Lemma 3.1 come from the asymptotic formulae (see, for example, [1, pp. 259, 260])

ψ(x)∼lnx− 1 2x −

∞

X

n=1

B_2n 2nx²ⁿ

= lnx− 1

2x − 1

12x² + 1

120x⁴ − 1

252x⁶ +· · · and

ψ⁰(x)∼ 1 x + 1

2x² +

∞

X

n=1

B_2n x²ⁿ⁺¹

= 1 x + 1

2x² + 1

6x³ − 1

30x⁵ + 1

42x⁷ − 1

30x⁹ +· · · , whereB_2nis the Bernoulli number of index2n.

Theorem 3.2. Leta ∈ (0,+∞). We consider the sequence (y_n)n∈N from the state- ment of Theorem2.1, the limit of which we denoted byγ(a).

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Then 1

2(a+n−1) +α ≤y_n−γ(a)< 1

2(a+n−1) +β, for eachn ∈N\ {1,2}, withα = _y ¹

3−γ(a) −2(a+ 2)andβ = ¹₃.

Moreover, the constantsαandβare the best possible with this property.

Proof. The inequalities from the statement of the theorem can be rewritten in the form

β < 1

y_n−γ(a) −2(a+n−1)≤α, for eachn∈N\ {1,2}.

Taking into account that ψ(x+ 1) = ψ(x) + ¹_x, for eachx ∈ (0,+∞), we can write that

ψ(a+n)−ψ(a) = 1 a + 1

a+ 1 +· · ·+ 1 a+n−1, for eachn∈N(see, for example, [1, p. 258]).

It is known that we have the series expansion (see, for example, [9, p. 336])

ψ(x) = lnx−

∞

X

k=0

1

x+k −ln

1 + 1 x+k

,

for eachx∈ (0,+∞). So, we are able to write the following relation betweenγ(a) and the logarithmic derivative of the gamma function:

γ(a) = lna−ψ(a) (see [6, Theorem 7], [11, Theorem 4.1, Remark 4.2]).

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Then

yn−γ(a) = ψ(a+n)−ψ(a)−lna+n−1

a −[lna−ψ(a)]

=ψ(a+n)−ln(a+n−1),

for eachn∈N. It means that, in fact, we have to prove that

β < 1

ψ(a+n)−ln(a+n−1) −2(a+n−1)≤α,

for eachn∈N\ {1,2}, and that the constantsαandβare the best possible with this property.

We consider the functionf : (0,+∞)→R, defined by f(x) = 1

ψ(x+ 1)−lnx −2x, for eachx∈(0,+∞). Differentiating, we get that

f⁰(x) =

1

x−ψ⁰(x+ 1)−2[ψ(x+ 1)−lnx]² [ψ(x+ 1)−lnx]² ,

for eachx∈(0,+∞). Using the inequalities from Lemma3.1, we are able to write that

1

x −ψ⁰(x+ 1)−2[ψ(x+ 1)−lnx]²

< 1

2x² − 1

6x³ + 1

30x⁵ − 1

42x⁷ + 1 30x⁹ −2

1

2x − 1

12x² + 1

120x⁴ − 1 252x⁶

2

=− 1

72x⁴ + 1

60x⁵ + 1

360x⁶ − 1

63x⁷ − 221

151200x⁸ + 1

30x⁹ + 1

7560x¹⁰ − 1 31752x¹²

=:g(x),

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for each x ∈ (0,+∞). It is not difficult to verify that g(x) < 0, for each x ∈ ₃

2,+∞

(³₂ not being the best lower value possible with this property). It follows thatf⁰(x) < 0, for eachx ∈ ₃

2,+∞

. So, the function f is strictly decreasing on ₃

2,+∞

. This means that the sequence (f(a +n −1))_n≥3 is strictly decreasing.

Therefore

k→∞lim f(a+k−1)< f(a+n−1)

≤f(a+ 2)

= 1

y₃−γ(a) −2(a+ 2), for eachn∈N\ {1,2}.

The asymptotic formula for the function ψ, mentioned in Remark4, permits us to write that

x→∞lim f(x) = lim

x→∞

1

6 +O _x¹2

1

2 +O ¹_x = 1 3.

Theorem 3.3. Leta ∈ ₁

2,+∞

. We consider the sequence(y_n)n∈Nfrom the state- ment of Theorem2.1, the limit of which we denoted byγ(a).

Then 1

2(a+n−1) +α ≤y_n−γ(a)< 1

2(a+n−1) +β, for eachn ∈N\ {1}, withα= _y ¹

2−γ(a)−2(a+ 1)andβ = ¹₃.

Proof. Sincea∈₁

2,+∞

, it follows that the sequence(f(a+n−1))n≥2 is strictly decreasing, wheref is the function defined in the proof of Theorem3.2.

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Theorem 3.4. Leta ∈ ₃

2,+∞

. We consider the sequence(y_n)n∈Nfrom the state- ment of Theorem2.1, the limit of which we denoted byγ(a).

Then 1

2(a+n−1) +α ≤y_n−γ(a)< 1

2(a+n−1) +β, for eachn ∈N, withα= _y ¹

1−γ(a) −2a = a[2aγ(a)−1]

1−aγ(a) andβ = ¹₃.

Proof. Sincea∈₃

2,+∞

, it follows that the sequence(f(a+n−1))n∈Nis strictly decreasing, wheref is the function defined in the proof of Theorem3.2.

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