Volume 9 (2008), Issue 2, Article 46, 7 pp.
SOME INEQUALITIES REGARDING A GENERALIZATION OF EULER’S
CONSTANT
ALINA SÎNTĂMĂRIAN
D EPARTMENT OF M ATHEMATICS
T ECHNICAL U NIVERSITY OF C LUJ -NAPOCA
S TR . C. DAICOVICIU NR . 15
400020 C LUJ -NAPOCA
ROMANIA .
Alina.Sintamarian@math.utcluj.ro
Received 28 November, 2007; accepted 20 March, 2008
Communicated by L. Tóth
A BSTRACT. The purpose of this paper is to evaluate the limit γ(a) of the sequence
1
1
a+n−1
1
+
+ ··· +
− ln
,
a a+1
a+n−1
a
n∈N
where a ∈ (0, +∞). We give some lower and upper estimates for
1
1
a+n−1
1
+
+ ··· +
− ln
− γ(a),
a a+1
a+n−1
a
n ∈ N.
Key words and phrases: Sequence, Convergence, Euler’s constant, Approximation, Estimate, Series.
2000 Mathematics Subject Classification. 11Y60, 40A05.
1. I NTRODUCTION
Let (Dn )n∈N be the sequence defined by Dn = 1 + 12 + · · · + n1 − ln n, for each n ∈ N. It
is well-known that the sequence (Dn )n∈N is convergent and its limit, usually denoted by γ, is
called Euler’s constant.
For Dn − γ, n ∈ N, many lower and upper estimates have been obtained in the literature. We
recall some of them:
1
1
< Dn − γ < 2(n−1)
, for each n ∈ N \ {1} ([14]);
• 2(n+1)
•
1
2(n+1)
•
1
2n+1
< Dn − γ <
1
,
2n
•
1
2n+ 25
< Dn − γ <
1
2n+ 13
•
1
2n+ 2γ−1
1−γ
352-07
< Dn − γ <
1
,
2n
≤ Dn − γ <
for each n ∈ N ([8], [19]);
for each n ∈ N ([17]);
, for each n ∈ N ([15], [16]);
1
2n+ 13
, for each n ∈ N ([16, Editorial comment], [2], [3]).
2
A LINA S ÎNTĂM ĂRIAN
In Section 2 we present a generalization of Euler’s constant as the limit of the sequence
1
1
1
a+n−1
, a ∈ (0, +∞),
+
+ ··· +
− ln
a a+1
a+n−1
a
n∈N
and we denote this limit by γ(a).
In Section 3 we give some lower and upper estimates for
1
1
1
a+n−1
+
+ ··· +
− ln
− γ(a),
a a+1
a+n−1
a
n ∈ N.
2. T HE N UMBER γ(a)
It is known that the sequence
1
1
a+n−1
1
+
+ ··· +
− ln
,
a a+1
a+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. Let a ∈ (0, +∞). We consider the sequences (xn )n∈N and (yn )n∈N defined by
1
1
1
a+n
xn = +
+ ··· +
− ln
a a+1
a+n−1
a
and
1
1
a+n−1
1
yn = +
+ ··· +
− ln
,
a a+1
a+n−1
a
for each n ∈ N.
Then:
(i) the sequences (xn )n∈N and (yn )n∈N are convergent to the same number, which we denote
by γ(a), and satisfy the inequalities xn < xn+1 < γ(a) < yn+1 < yn , for each n ∈ N;
(ii) 0 < a1 − ln 1 + a1 < γ(a) < a1 ;
(iii) lim n(γ(a) − xn ) =
n→∞
1
2
and lim n(yn − γ(a)) = 12 .
n→∞
Remark 1. The sequence (yn )n∈N from Theorem 2.1, for a = 1, becomes the sequence (Dn )n∈N ,
so γ(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
, for each n ∈ N, where (yn )n∈N is the sequence from the statement of Theorem 2.1.
2(a+n−1)+ 31
Also, we specify that γ(a) is the limit of the sequence (yn )n∈N .
Then:
1
(i) un < un+1 < γ(a), for each n ∈ N \ {1}, and lim n3 (γ(a) − un ) = 72
;
n→∞
(ii)
1
2(a+n−1)+ 11
28
< yn − γ(a) <
1
2(a+n−1)+ 13
, for each n ∈ N \ {1}.
Remark 2. The lower estimate from part (ii) of Theorem 2.2 holds for n = 1 as well.
Remark 3. The second limit from part (iii) of Theorem 2.1 also follows from part (ii) of
Theorem 2.2.
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 46, 7 pp.
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G ENERALIZATION OF E ULER ’ S C ONSTANT
3
3. P ROVING S OME E STIMATES FOR yn − γ(a) USING THE L OGARITHMIC D ERIVATIVE
OF THE G AMMA F UNCTION
As we already mentioned in Section 1, it is known that ([16, Editorial comment], [2, Theorem
3], [3, Theorem 1.1])
1
1
,
2γ−1 ≤ Dn − γ <
2n + 13
2n + 1−γ
for each n ∈ N, the constants 2γ−1
and 13 being the best possible with this property.
1−γ
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 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 by γ(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) − ln x >
(ii)
1
x
− ψ 0 (x + 1) <
1
2x
1
2x2
−
−
1
12x2
1
6x3
+
+
1
120x4
1
30x5
−
−
1
,
252x6
1
42x7
+
for each x ∈ (0, +∞);
1
,
30x9
for each x ∈ (0, +∞).
We specify that the function ψ is the logarithmic derivative of the gamma function, i.e. ψ(x) =
Γ0 (x)
, for each x ∈ (0, +∞).
Γ(x)
Proof. (i) It is known (see, for example, [18, p. 116]) that ln x =
x ∈ (0, +∞). Also, we shall need the formula
Z ∞ −t
e
e−xt
ψ(x) =
−
dt,
t
1 − e−t
0
R∞
0
e−t −e−xt
t
dt, for each
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
Z ∞
1
1
ψ(x + 1) − ln x =
− t
e−xt dt,
t
e
−
1
0
for each x ∈ (0, +∞).
It is not difficult to verify that
Z
∞
0
tn e−xt dt =
n!
,
xn+1
for each n ∈ N ∪ {0}, any x ∈ (0, +∞).
Then we have
1
1
1
1
ψ(x + 1) − ln x −
+
−
+
2x 12x2 120x4 252x6
Z ∞
1
1
1
t
t3
t5
=
− t
− +
−
+
e−xt dt
t
e
−
1
2
12
720
30240
Z0 ∞
1
=
[30240(et − 1) − 30240t − 15120t(et − 1) + 2520t2 (et − 1)
t − 1)
30240t(e
0
− 42t4 (et − 1) + t6 (et − 1)]e−xt dt
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 46, 7 pp.
http://jipam.vu.edu.au/
4
A LINA S ÎNTĂM ĂRIAN
Z
∞
=
0
Z
=
∞
"
∞
∞ n+1
∞ n+2
X
X
X
1
tn
t
t
30240
− 15120
+ 2520
t
30240t(e − 1)
n!
n!
n!
n=2
n=1
n=1
#
∞ n+6
∞ n+4
X
X
t
t
−42
+
e−xt dt
n!
n!
n=1
n=1
∞
P
n=9
(n−3)(n−5)(n−7)(n−8)(n2 +8n+36) n
t
n!
30240t(et − 1)
0
· e−xt dt > 0,
for each x ∈ (0, +∞).
(ii) In part (i) we obtained that
Z
ln x − ψ(x + 1) =
0
∞
1
1
−
et − 1 t
e−xt dt,
for each x ∈ (0, +∞). Differentiating here we get that
Z ∞
1
t
0
− ψ (x + 1) =
1− t
e−xt dt,
x
e
−
1
0
for each x ∈ (0, +∞).
Then we have
1
1
1
1
1
1
− ψ 0 (x + 1) − 2 + 3 −
+
−
x Z 2x
6x
30x5 42x7 30x9
∞
2
t
t
t4
t6
t8
t
− +
−
+
−
e−xt dt
=
1− t
e
−
1
2
12
720
30240
1209600
Z0 ∞
1
[1209600(et − 1) − 1209600t − 604800t(et − 1)
=
t − 1)
1209600(e
0
+ 100800t2 (et − 1) − 1680t4 (et − 1) + 40t6 (et − 1) − t8 (et − 1)]e−xt dt
"
Z ∞
∞
∞ n+1
X
X
1
tn
t
=
1209600
− 604800
t
1209600(e − 1)
n!
n!
0
n=2
n=1
#
∞ n+2
∞ n+4
∞ n+6
∞ n+8
X
X
X
X
t
t
t
t
+100800
− 1680
+ 40
−
e−xt dt
n!
n!
n!
n!
n=1
n=1
n=1
n=1
Z ∞ P∞ (n−3)(n−5)(n−7)(n−9)(n−10)(n+4)(n2 +2n+32) n
t
n=11
n!
=−
· e−xt dt < 0,
t
1209600(e
−
1)
0
for each x ∈ (0, +∞).
Remark 4. In fact, these inequalities from Lemma 3.1 come from the asymptotic formulae
(see, for example, [1, pp. 259, 260])
∞
X B2n
1
ψ(x) ∼ ln x −
−
2x n=1 2nx2n
= ln x −
1
1
1
1
−
+
−
+ ···
2
4
2x 12x
120x
252x6
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 46, 7 pp.
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G ENERALIZATION OF E ULER ’ S C ONSTANT
5
and
∞
X B2n
1
1
ψ (x) ∼ + 2 +
x 2x
x2n+1
n=1
0
1
1
1
1
1
1
+ 2+ 3−
+
−
+ ··· ,
x 2x
6x
30x5 42x7 30x9
where B2n is the Bernoulli number of index 2n.
=
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 by γ(a).
Then
1
1
≤ yn − γ(a) <
,
2(a + n − 1) + α
2(a + n − 1) + β
1
for each n ∈ N \ {1, 2}, with α = y3 −γ(a)
− 2(a + 2) and β = 31 .
Moreover, the constants α and β are the best possible with this property.
Proof. The inequalities from the statement of the theorem can be rewritten in the form
1
β<
− 2(a + n − 1) ≤ α,
yn − γ(a)
for each n ∈ N \ {1, 2}.
Taking into account that ψ(x + 1) = ψ(x) + x1 , for each x ∈ (0, +∞), we can write that
1
1
1
ψ(a + n) − ψ(a) = +
+ ··· +
,
a a+1
a+n−1
for each n ∈ N (see, for example, [1, p. 258]).
It is known that we have the series expansion (see, for example, [9, p. 336])
∞ X
1
1
ψ(x) = ln x −
− ln 1 +
,
x+k
x+k
k=0
for each x ∈ (0, +∞). So, we are able to write the following relation between γ(a) and the
logarithmic derivative of the gamma function:
γ(a) = ln a − ψ(a)
(see [6, Theorem 7], [11, Theorem 4.1, Remark 4.2]).
Then
a+n−1
yn − γ(a) = ψ(a + n) − ψ(a) − ln
− [ln a − ψ(a)]
a
= ψ(a + n) − ln(a + n − 1),
for each n ∈ N. It means that, in fact, we have to prove that
1
β<
− 2(a + n − 1) ≤ α,
ψ(a + n) − ln(a + n − 1)
for each n ∈ N \ {1, 2}, and that the constants α and β are the best possible with this property.
We consider the function f : (0, +∞) → R, defined by
1
f (x) =
− 2x,
ψ(x + 1) − ln x
for each x ∈ (0, +∞). Differentiating, we get that
f 0 (x) =
1
x
− ψ 0 (x + 1) − 2[ψ(x + 1) − ln x]2
,
[ψ(x + 1) − ln x]2
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 46, 7 pp.
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6
A LINA S ÎNTĂM ĂRIAN
for each x ∈ (0, +∞). Using the inequalities from Lemma 3.1, we are able to write that
1
− ψ 0 (x + 1) − 2[ψ(x + 1) − ln x]2
x
2
1
1
1
1
1
1
1
1
1
< 2− 3+
−
+
−2
−
+
−
2x
6x
30x5 42x7 30x9
2x 12x2 120x4 252x6
1
1
1
221
1
1
1
1
=−
+
+
−
−
+
+
−
4
5
6
7
8
9
10
72x
60x
360x
63x
151200x
30x
7560x
31752x12
=: g(x),
for each x ∈ (0, +∞). It is not difficult to verify that g(x) < 0, for each x ∈ 23 , +∞
0
( 23 not being
3 the best
lower value possible with this property). It3 follows
that f (x) < 0, for
each x ∈ 2 , +∞ . So, the function f is strictly decreasing on 2 , +∞ . This means that the
sequence (f (a + n − 1))n≥3 is strictly decreasing. Therefore
lim f (a + k − 1) < f (a + n − 1)
k→∞
≤ f (a + 2)
1
=
− 2(a + 2),
y3 − γ(a)
for each n ∈ N \ {1, 2}.
The asymptotic formula for the function ψ, mentioned in Remark 4, permits us to write that
1
+ O x12
1
6
= .
lim f (x) = lim 1
1
x→∞
x→∞
3
+O x
2
1
Theorem 3.3. Let a ∈ 2 , +∞ . We consider the sequence (yn )n∈N from the statement of
Theorem 2.1, the limit of which we denoted by γ(a).
Then
1
1
≤ yn − γ(a) <
,
2(a + n − 1) + α
2(a + n − 1) + β
1
for each n ∈ N \ {1}, with α = y2 −γ(a)
− 2(a + 1) and β = 31 .
Moreover, the constants α and β are the best possible with this property.
Proof. Since a ∈ 12 , +∞ , it follows that the sequence (f (a + n − 1))n≥2 is strictly decreasing,
where f is the function defined in the proof of Theorem 3.2.
Theorem 3.4. Let a ∈ 32 , +∞ . We consider the sequence (yn )n∈N from the statement of
Theorem 2.1, the limit of which we denoted by γ(a).
Then
1
1
≤ yn − γ(a) <
,
2(a + n − 1) + α
2(a + n − 1) + β
1
for each n ∈ N, with α = y1 −γ(a)
− 2a = a[2aγ(a)−1]
and β = 13 .
1−aγ(a)
Moreover, the constants α and β are the best possible with this property.
Proof. Since a ∈ 32 , +∞ , it follows that the sequence (f (a + n − 1))n∈N is strictly decreasing,
where f is the function defined in the proof of Theorem 3.2.
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 46, 7 pp.
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G ENERALIZATION OF E ULER ’ S C ONSTANT
7
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J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 46, 7 pp.
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