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SOME INEQUALITIES REGARDING A GENERALIZATION OF EULER’S CONSTANT

ALINA SÎNT ˘AM ˘ARIAN DEPARTMENT OFMATHEMATICS

TECHNICALUNIVERSITY OFCLUJ-NAPOCA

STR. C. DAICOVICIU NR. 15 400020 CLUJ-NAPOCA

ROMANIA.

Alina.Sintamarian@math.utcluj.ro

Received 28 November, 2007; accepted 20 March, 2008 Communicated by L. Tóth

ABSTRACT. The purpose of this paper is to evaluate the limitγ(a)of the sequence

1

a+ 1

a+ 1 +· · ·+ 1

a+n1lna+n1 a

n∈N

,

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

a+ 1

a+ 1 +· · ·+ 1

a+n1lna+n1

a γ(a), nN.

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

2000 Mathematics Subject Classification. 11Y60, 40A05.

1. INTRODUCTION

Let(Dn)n∈Nbe the sequence defined by Dn = 1 + 12 +· · ·+ 1n −lnn, for eachn ∈ N. It is well-known that the sequence (Dn)n∈Nis convergent and its limit, usually denoted by γ, 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 < Dn−γ < 2(n−1)1 , for eachn∈N\ {1}([14]);

2(n+1)1 < Dn−γ < 2n1 , for eachn∈N([8], [19]);

2n+11 < Dn−γ < 2n1 , for eachn∈N([17]);

2n+1 2 5

< Dn−γ < 2n+1 1 3

, for eachn ∈N([15], [16]);

1

2n+2γ−11−γ ≤Dn−γ < 1

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

352-07

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In Section 2 we present a generalization of Euler’s constant as the limit of the sequence 1

a + 1

a+ 1 +· · ·+ 1

a+n−1−lna+n−1 a

n∈N

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

In Section 3 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.

2. THENUMBERγ(a) It is known that the sequence

1 a + 1

a+ 1 +· · ·+ 1

a+n−1−lna+n−1 a

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(xn)n∈Nand(yn)n∈Ndefined by xn = 1

a + 1

a+ 1 +· · ·+ 1

a+n−1 −lna+n a and

yn= 1 a + 1

a+ 1 +· · ·+ 1

a+n−1−lna+n−1

a ,

for eachn ∈N. Then:

(i) the sequences(xn)n∈Nand(yn)n∈Nare convergent to the same number, which we denote byγ(a), and satisfy the inequalitiesxn< xn+1 < γ(a)< yn+1 < yn, for eachn ∈N; (ii) 0< a1 −ln 1 + 1a

< γ(a)< 1a; (iii) lim

n→∞n(γ(a)−xn) = 12 and lim

n→∞n(yn−γ(a)) = 12.

Remark 1. The sequence(yn)n∈Nfrom Theorem 2.1, fora= 1, becomes the sequence(Dn)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(un)n∈N defined by un = yn

1

2(a+n−1)+13, for eachn ∈N, where(yn)n∈Nis the sequence from the statement of Theorem 2.1.

Also, we specify thatγ(a)is the limit of the sequence(yn)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

< yn−γ(a)< 2(a+n−1)+1 1 3

, for eachn∈N\ {1}.

Remark 2. The lower estimate from part(ii)of Theorem 2.2 holds forn= 1as well.

Remark 3. The second limit from part (iii) of Theorem 2.1 also follows from part (ii) of Theorem 2.2.

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3. PROVINGSOMEESTIMATES FORyn−γ(a)USING THE LOGARITHMICDERIVATIVE OF THEGAMMA FUNCTION

As we already mentioned in Section 1, it is known that ([16, Editorial comment], [2, Theorem 3], [3, Theorem 1.1])

1

2n+ 2γ−11−γ ≤Dn−γ < 1 2n+13,

for eachn∈N, the constants 2γ−11−γ and 13 being the best possible with this property.

Leta ∈ (0,+∞). In a similar way as in the proof given by H. Alzer in [2, Theorem 3], we shall obtain lower and upper estimates for yn−γ(a)(n ∈ N), where (yn)n∈N is the sequence from the statement of Theorem 2.1, 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 > 2x112x12 + 120x1 4252x1 6, for eachx∈(0,+∞);

(ii) 1x −ψ0(x+ 1)< 2x126x13 + 30x1542x17 + 30x19, for eachx∈(0,+∞).

We specify that the functionψ is the logarithmic derivative of the gamma function, i.e. ψ(x) =

Γ0(x)

Γ(x), for eachx∈(0,+∞).

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

e−t−e−xt

t dt, for each x∈(0,+∞). Also, we shall need the formula

ψ(x) = Z

0

e−t

t − e−xt 1−e−t

dt,

which holds for eachx∈ (0,+∞), known as Gauss’ expression ofψ(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 et−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

et−1− 1 2 + t

12− t3

720 + t5 30240

e−xtdt

= Z

0

1

30240t(et−1)[30240(et−1)−30240t−15120t(et−1) + 2520t2(et−1)

−42t4(et−1) +t6(et−1)]e−xtdt

(4)

= Z

0

1 30240t(et−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

= Z

0

P

n=9

(n−3)(n−5)(n−7)(n−8)(n2+8n+36)

n! tn

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

(ii)In part(i)we obtained that

lnx−ψ(x+ 1) = Z

0

1

et−1− 1 t

e−xtdt,

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

x −ψ0(x+ 1) = Z

0

1− t et−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

et−1 − t 2 + t2

12 − t4

720 + t6

30240 − t8 1209600

e−xtdt

= Z

0

1

1209600(et−1)[1209600(et−1)−1209600t−604800t(et−1)

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

= Z

0

1

1209600(et−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(et−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

B2n

2nx2n

= lnx− 1

2x − 1

12x2 + 1

120x4 − 1

252x6 +· · ·

(5)

and

ψ0(x)∼ 1 x + 1

2x2 +

X

n=1

B2n x2n+1

= 1 x + 1

2x2 + 1

6x3 − 1

30x5 + 1

42x7 − 1

30x9 +· · · , whereB2nis the Bernoulli number of index2n.

Theorem 3.2. Let a ∈ (0,+∞). We consider the sequence (yn)n∈N from the statement of Theorem 2.1, the limit of which we denoted byγ(a).

Then 1

2(a+n−1) +α ≤yn−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

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

Taking into account thatψ(x+ 1) =ψ(x) + 1x, 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]).

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 ,

(6)

for eachx∈(0,+∞). Using the inequalities from Lemma 3.1, 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),

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 that f0(x) < 0, for eachx ∈ 3

2,+∞

. So, the functionf is strictly decreasing on3

2,+∞

. This means that the sequence(f(a+n−1))n≥3is 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 functionψ, mentioned in Remark 4, permits us to write that

x→∞lim f(x) = lim

x→∞

1

6 +O x12

1

2 +O 1x = 1 3.

Theorem 3.3. Let a ∈ 1

2,+∞

. We consider the sequence (yn)n∈N from the statement of Theorem 2.1, the limit of which we denoted byγ(a).

Then

1

2(a+n−1) +α ≤yn−γ(a)< 1

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

2−γ(a) −2(a+ 1)andβ = 13.

Moreover, the constantsαandβare the best possible with this property.

Proof. Sincea∈1

2,+∞

, it follows that the sequence(f(a+n−1))n≥2is strictly decreasing,

wheref is the function defined in the proof of Theorem 3.2.

Theorem 3.4. Let a ∈ 3

2,+∞

. We consider the sequence (yn)n∈N from the statement of Theorem 2.1, the limit of which we denoted byγ(a).

Then

1

2(a+n−1) +α ≤yn−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.

Moreover, the constantsαandβare the best possible with this property.

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 Theorem 3.2.

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REFERENCES

[1] M. ABRAMOWITZ AND I.A. STEGUN, Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, National Bureau of Standards Applied Mathematics Series 55, Washington, 1964.

[2] H. ALZER, Inequalities for the gamma and polygamma functions, Abh. Math. Sem. Univ. Hamburg, 68 (1998), 363–372.

[3] C.-P. CHEN AND F. QI, The best lower and upper bounds of harmonic sequence, RGMIA 6(2) (2003), 303–308.

[4] D.W. DeTEMPLE, A quicker convergence to Euler’s constant, Amer. Math. Monthly, 100(5) (1993), 468–470.

[5] K. KNOPP, Theory and Application of Infinite Series, Blackie & Son Limited, London and Glas- gow, 1951.

[6] D.H. LEHMER, Euler constants for arithmetical progressions, Acta Arith., 27 (1975), 125–142.

[7] I. NEDELCU, Problem 21753, Gazeta Matematic˘a, Seria B, 94(4) (1989), 136.

[8] P.J. RIPPON, Convergence with pictures, Amer. Math. Monthly, 93(6) (1986), 476–478.

[9] I.M. RÎJICANDI.S. GRAD ¸STEIN, Tabele de integrale. Sume, serii ¸si produse (Tables of Integrals.

Sums, Series and Products), Editura Tehnic˘a, Bucure¸sti, 1955.

[10] A. SÎNT ˘AM ˘ARIAN, Approximations for a generalization of Euler’s constant (submitted).

[11] A. SÎNT ˘AM ˘ARIAN, About a generalization of Euler’s constant, Aut. Comp. Appl. Math., 16(1) (2007), 153–163.

[12] A. SÎNT ˘AM ˘ARIAN, A generalization of Euler’s constant, Numer. Algorithms, 46(2) (2007), 141–

151.

[13] T. TASAKA, Note on the generalized Euler constants, Math. J. Okayama Univ., 36 (1994), 29–34.

[14] S.R. TIMSANDJ.A. TYRRELL, Approximate evaluation of Euler’s constant, Math. Gaz., 55(391) (1971), 65–67.

[15] L. TÓTH, Problem E3432, Amer. Math. Monthly, 98(3) (1991), 264.

[16] L. TÓTH, Problem E3432 (Solution), Amer. Math. Monthly, 99(7) (1992), 684–685.

[17] A. VERNESCU, Ordinul de convergen¸t˘a al ¸sirului de defini¸tie al constantei lui Euler (The conver- gence order of the definition sequence of Euler’s constant), Gazeta Matematic˘a, Seria B, 88(10-11) (1983), 380–381.

[18] E.T. WHITTAKER AND G.N. WATSON, A Course of Modern Analysis, Cambridge University Press, Cambridge, 1996.

[19] R.M. YOUNG, Euler’s constant, Math. Gaz., 75(472) (1991), 187–190.

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