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```CUBO A Mathematial Journal
Vol.15, No 03, (105122). Otober 2013
Euler's onstant, new lasses of sequenes and estimates
Alina Sînt m rian
Department of Mathematis,
Tehnial University of Cluj-Napoa,
Str. Memorandumului nr. 28, 400114 Cluj-Napoa, Romania.
alina.sintamarianmath.utluj.ro
ABSTRACT
We give two lasses of sequenes with the argument of the logarithmi term modied and also with some additional terms besides those in the denition sequene, and
Pn
1
− ln a+n−1
that onverge quikly to γ(a) = lim
, where a ∈ (0, +∞).
k=1
a+k−1
a
n→∞
We present the pattern in forming these sequenes, expressing the oeients that
appear with the Bernoulli numbers. Also, we obtain estimates ontaining best onPn 1
1
1
7
stants for
γ−
k=1 k − 24(n+1/2)2 − ln n + 2 − 960(n+1/2)3 − γ and
P
n
7
1
1
1
31
, where γ = γ(1)
k=1 k − 24(n+1/2)2 + 960(n+1/2)4 − ln n + 2 + 8064(n+1/2)5
is the Euler's onstant.
RESUMEN
Mostramos dos lases de seuenias on el argumento del término logarítmio modiado y también on algunos términos adiionales además de los denidos en la seuen
Pn
1
a+n−1
ia y que onvergen rápidamente a γ(a) = lim
, donde a ∈
k=1 a+k−1 − ln
a
n→∞
(0, +∞). Presentamos el patrón que forma las seuenias expresando los oeientes
que apareen en los números de Bernoulli. Además, obtenemos
estimaiones que onPn 1
1
7
1
tienen las mejores onstantes para
k=1 k − 24(n+1/2)2 − ln n + 2 − 960(n+1/2)3 −γ
P
n
7
1
1
31
1
y γ−
, donde γ =
k=1 k − 24(n+1/2)2 + 960(n+1/2) 4 − ln n + 2 + 8064(n+1/2)5
γ(1) es la onstante de Euler.
Keywords and Phrases:
sequene, onvergene, approximation, Euler's onstant, Bernoulli
number, estimate.
2010 AMS Mathematis Subjet Classiation:
11Y60, 11B68, 40A05, 41A44, 33B15.
106
1
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Alina Sînt m rian
15, 3 (2013)
Introdution
Pn
Let Hn = k=1 1/k be the nth harmoni number and let Dn = Hn − ln n. Euler's onstant
γ = limn→∞ Dn is one of the most important onstants in mathematis and is also the topi of
many papers in the literature. This omes as a onrmation of what Leonhard Euler said about
γ, namely that it is worthy of serious onsideration ( [10, pp. xx, 51℄).
It is well-known (see [19℄, [20℄, [2℄, [5℄) that
1
1
≤ Dn − γ <
2γ−1
2n +
2n + 1−γ
1
3
,
n ∈ N,
2γ−1
1
the numbers 2γ−1
1−γ and 3 being the best onstants with this property, i.e. 1−γ annot be replaed
1
by a smaller one and 3 annot be replaed by a larger one, so that the above-mentioned inequalities
to hold for all n ∈ N. Having in view that limn→∞ n(Dn − γ) = 1/2, one an say that the sequene
(Dn )n∈N onverges to γ very slowly. In order to inrease the speed of onvergene to γ, D. W.
DeTemple [7℄ modied the argument of the logarithmi term from Dn , onsidering the sequene
1
1
(Rn )n∈N dened by Rn = Hn − ln(n + 1/2), and he proved that 24(n+1)
2 < Rn − γ < 24n2 , n ∈ N.
Sequenes with higher rate of onvergene to γ an be also obtained by subtrating a rational term
from Dn : it is shown (see [21℄) that limn→∞ n2 (γ−Dn +1/(2n)) = 1/12. In our paper we shall try
to ombine these two methods, modifying the argument of the logarithmi term, and subtrating
and adding terms in the denition sequene, to obtain quiker onvergenes to a generalization
of Euler's onstant. Also, we shall provide estimates regarding Euler's onstant γ and this is the
reason why further on we remind some of the estimates related to γ and ontaining best onstants,
that have been obtained in the literature:
1
24(n+a1 )2
≤ Rn − γ <
1
,
24(n+b1 )2
n∈N
([3℄);
1
([9℄);
≤ 12n21+b2 , n ∈ N
< γ − Dn − 2n
7
1
1
7
n∈N
< 960(n+b
4,
960(n+a3 )4 ≤ γ − Hn − ln n + 2 − 24(n+1/2)2
3)
17
23
17
1
1
− γ < 3840(n+b
≤ Hn − ln n + 12 + 24n
− 48n
2 + 5760n3
5,
3840(n+a4 )5
4)
1
12n2 +a2
([4℄);
n∈N
([6℄),
with
a1 = 1/
p
24 (1 − γ − ln(3/2)) − 1 and b1 = 1/2;
a2 = 6/5 and b2 = 2(7 − 12γ)/(2γ − 1);
p
a3 = 1/ 4 960/7 (ln(3/2) + γ − 53/54) − 1 and b3 = 1/2;
p
a4 = 1/ 5 3840/17 (1 − γ − ln(8783/5760)) − 1 and b4 = 3305/12852,
where ai and bi are the best onstants with the property that the orresponding inequalities hold
for all n ∈ N, i ∈ {1, 2, 3, 4}.
As we antiipated earlier, in the present paper we shall investigate a generalization of Euler's
onstant, namely the limit γ(a) of the sequene (yn (a))n∈N dened by (see [11, p. 453℄, [16℄, [17℄,
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15, 3 (2013)
Euler's onstant, new lasses of sequenes and estimates
[18℄)
yn (a) =
n
X
107
a+n−1
1
− ln
,
a+k−1
a
k=1
where a ∈ (0, +∞). Obviously, γ(1) = γ. Numerous results related to γ(a) an be found, for
example, in [16℄, [17℄, [18℄, [14℄, [12℄.
In Setion 2 we give two lasses of sequenes with the argument of the logarithmi term
modied and also with some additional terms besides those in the denition sequene (yn (a))n∈N ,
and that onverge quikly to γ(a). We present the pattern in forming these sequenes, expressing
the oeients that appear with the Bernoulli numbers. The two lasses of sequenes have as base
the sequenes (αn,2 (a))n∈N and (αn,3 (a))n∈N dened by
!
n
X
a + n − 12
1
1
7
αn,2 (a) =
−
−
− ln
3 ,
a + k − 1 24 a + n − 1 2
a
960a a + n − 1
αn,3 (a) =
k=1
n
X
2
2
1
1
7
−
+
a + k − 1 24 a + n − 1 2 960 a + n − 1 4
k=1
2
2
!
1
a+n− 2
31
− ln
+
5 .
a
8064a a + n − 1
2
In Setion 3 we prove estimates ontaining best onstants for αn,2 (1) − γ and γ − αn,3 (1), n ∈ N.
The following lemma, whih we shall need in our proofs, was given by C. Mortii [13, Lemma℄
and is a onsequene of the the Stolz-Cesàro Theorem, the 0/0 ase [8, Theorem B.2, p. 265℄.
Lemma 1.1.
Let (xn )n∈N be a onvergent sequene of real numbers and x∗ = lim xn . We suppose
n→∞
that there exists α ∈ R, α > 1, suh that
lim nα (xn − xn+1 ) = l ∈ R.
n→∞
Then there exists the limit
lim nα−1 (xn − x∗ ) =
n→∞
l
.
α−1
Also, reall that the digamma funtion ψ is the logarithmi derivative of the gamma funtion,
i.e.
ψ(x) =
Γ ′ (x)
,
Γ (x)
x ∈ (0, +∞).
It is known that ( [1, p. 258℄, [15, p. 337℄)
ψ(n + 1) = −γ + Hn ,
n ∈ N.
From the reurrene formula ( [1, p. 258℄)
1
ψ(x + 1) = ψ(x) + ,
x
x ∈ (0, +∞),
(1)
108
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Alina Sînt m rian
15, 3 (2013)
and the asymptoti formula ( [1, p. 259℄)
ψ(x) ∼ ln x −
1
1
1
1
1
1
+
−
+
−
+ ···
−
2x 12x2
120x4
252x6
240x8
132x10
(x → ∞),
one obtains
ψ(x + 1) ∼ ln x +
2
1
1
1
1
1
1
+
−
+
−
+ ···
−
2
4
6
8
2x 12x
120x
252x
240x
132x10
Sequenes that onverge to
(x → ∞).
(2)
γ(a)
Theorem 2.1.
Let a ∈ (0, +∞) and let γ(a) be the limit of the sequene (yn (a))n∈N from Introdution. We onsider the sequenes (αn,2 (a))n∈N and (βn,2 (a))n∈N dened by
!
n
X
a + n − 12
1
1
7
αn,2 (a) =
−
−
− ln
3 ,
a + k − 1 24 a + n − 1 2
a
960a a + n − 1
k=1
2
βn,2 (a) = αn,2 (a) −
31
8064 a + n −
Then:
(i) lim n6 (αn,2 (a) − γ(a)) =
n→∞
(ii) lim n8 (γ(a) − βn,2 (a)) =
n→∞
Proof. (i) Set εn :=
1
2
εn −
7
960
we obtain
·
1
a+n ,
ε4
n
3
1+ 1
2 εn
(
)
1 6
2
2
.
31
;
8064
7571
.
1843200
n ∈ N. Sine ± 12 εn ∈ (−1, 1), − 21 εn −
7
960
·
ε4
n
(1− 12 εn )
3
∈ (−1, 1] and
∈ (−1, 1], for every n ∈ N, using the series expansion ( [11, pp. 171179℄)
αn,2 (a) − αn+1,2 (a)
1
ε2n
1
ε2n
= −εn −
·
+
·
24 1 − 1 εn 2
24 1 + 1 εn 2
2
!2
!
4
1
1
7
εn
7
ε4n
− ln 1 − εn −
+ ln 1 + εn −
·
·
2
960 1 − 1 εn 3
2
960 1 + 1 εn 3
2
2
4829 9
2913 11
20456239 13
31 7
15
ε +
ε +
ε +
ε + O(εn ).
=
1344 n 230400 n 225280 n
2875392000 n
It follows that
lim n7 (αn,2 (a) − αn+1,2 (a)) =
n→∞
31
.
1344
Now, aording to Lemma 1.1, we get
lim n6 (αn,2 (a) − γ(a)) =
n→∞
31
.
8064
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Euler's onstant, new lasses of sequenes and estimates
15, 3 (2013)
109
(ii) We are able to write that
βn+1,2 (a) − βn,2 (a)
= αn+1,2 (a) − αn,2 (a) −
=
ε6n
ε6n
31
31
·
·
+
1
8064 (1 + 2 εn )6 8064 (1 − 21 εn )6
10727 11
7571 9
ε +
ε + O(ε13
n ).
230400 n 225280 n
Therefore
lim n9 (βn+1,2 (a) − βn,2 (a)) =
n→∞
7571
,
230400
and based on Lemma 1.1 we obtain
lim n8 (γ(a) − βn,2 (a)) =
n→∞
7571
.
1843200
Also, onsidering the sequene in eah of the following parts and using similar arguments as
in Theorem 2.1, we get the indiated limit:
7571
δn,2 (a) = βn,2 (a) +
8 , n ∈ N,
1843200 a + n − 21
lim n10 (δn,2 (a) − γ(a)) =
n→∞
ηn,2 (a) = δn,2 (a) −
511
67584 a + n −
1 10
2
, n ∈ N,
lim n12 (γ(a) − ηn,2 (a)) =
n→∞
θn,2 (a) = ηn,2 (a) +
5092085987
241532928000 a + n −
1 12
2
5092085987
;
241532928000
, n ∈ N,
lim n14 (θn,2 (a) − γ(a)) =
n→∞
λn,2 (a) = θn,2 (a) −
8191
98304 a + n −
1 14
2
lim n16 (γ(a) − λn,2 (a)) =
25599939583183
;
57755566080000
25599939583183
57755566080000 a + n −
1 16
2
lim n18 (µn,2 (a) − γ(a)) =
n→∞
8191
;
98304
, n ∈ N,
n→∞
µn,2 (a) = λn,2 (a) +
511
;
67584
, n ∈ N,
5749691557
.
1882718208
110
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Alina Sînt m rian
15, 3 (2013)
We remark the pattern in forming the sequenes from Theorem 2.1 and those mentioned above.
For example, the general term of the sequene (µn,2 (a))n∈N an be written in the form
µn,2 (a) =
n
X
1
1 B2
1
− ·
·
2
a+k−1 2 2
a + n − 12
k=1
− ln
with
ck,2
a+n−
a
 2k−1
2
−1



22k−1
=

22k−1 − 1


22k−1
1
2
23 − 1 B4
1
·
+
·
3
2
4 a a + n − 1 3
2
!
−
8
X
k=3
ck,2
a+n−
1 2k
2
,
B2k
if k = 2p + 1, p ∈ N,
,
2k
3
k2
B2k
2 − 1 B4
2
− 3 ·
·
+
, if k = 2p + 2, p ∈ N,
2k
k
2
4
·
where B2k is the 2kth Bernoulli number. Related to this remark, see also [16, Remark 3.4℄, [18, p.
71, Remark 2.1.3; pp. 100, 101, Remark 3.1.6℄.
For Euler's onstant γ = 0.5772156649 . . . we obtain, for example:
α2,2 (1) = 0.5772292855 . . . ;
α3,2 (1) = 0.5772175963 . . . ;
β2,2 (1) = 0.5772135395 . . . ;
β3,2 (1) = 0.5772155051 . . . ;
δ2,2 (1) = 0.5772162314 . . . ;
δ3,2 (1) = 0.5772156875 . . . ;
η2,2 (1) = 0.5772154386 . . . ;
η3,2 (1) = 0.5772156600 . . . ;
θ2,2 (1) = 0.5772157923 . . . ;
θ3,2 (1) = 0.5772156663 . . . ;
λ2,2 (1) = 0.5772155686 . . . ;
λ3,2 (1) = 0.5772156643 . . . ;
µ2,2 (1) = 0.5772157590 . . . ;
µ3,2 (1) = 0.5772156651 . . . .
As an be seen, λ3,2 (1) is aurate to nine deimal plaes in approximating γ.
Theorem 2.2. Let a ∈ (0, +∞) and let γ(a) be the limit of the sequene (yn (a))n∈N from Introdution. We onsider the sequenes (αn,3 (a))n∈N , (βn,3 (a))n∈N and (δn,3 (a))n∈N dened by
αn,3 (a) =
n
X
7
1
1
+
−
4
a + k − 1 24 a + n − 1 2
960 a + n − 12
k=1
2
!
a + n − 12
31
− ln
+
5 ,
a
8064a a + n − 1
2
βn,3 (a) = αn,3 (a) +
δn,3 (a) = βn,3 (a) −
127
30720 a + n −
511
67584 a + n −
Then:
(i) lim n8 (γ(a) − αn,3 (a)) =
n→∞
127
;
30720
1 8
2
,
1 10
2
.
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Euler's onstant, new lasses of sequenes and estimates
15, 3 (2013)
(ii) lim n10 (βn,3 (a) − γ(a)) =
n→∞
(iii) lim n12 (γ(a) − δn,3 (a)) =
n→∞
Proof. (i) Set εn :=
31
·
− 21 εn + 8064
we obtain
1
a+n ,
ε6
n
5
1− 1
2 εn
(
)
111
511
;
67584
178161637
.
8453652480
n ∈ N. Sine ± 21 εn ∈ (−1, 1),
1
2
εn +
31
8064
·
ε6
n
5
(1+ 12 εn )
∈ (−1, 1] and
∈ (−1, 1], for every n ∈ N, using the series expansion ( [11, pp. 171179℄)
αn+1,3 (a) − αn,3 (a)
1
7
7
ε2n
ε2n
ε4n
ε4n
1
+
+
−
·
·
·
·
= εn −
2
2
4
24 1 + 1 εn
24 1 − 1 εn
960 1 + 1 εn
960 1 − 1 εn 4
2
2
2
2
!
!
31
ε6n
31
ε6n
1
1
+ ln 1 − εn +
− ln 1 + εn +
·
·
2
8064 1 + 1 εn 5
2
8064 1 − 1 εn 5
2
2
127 9
409 11
5873471 13
2502391 15
=
ε +
ε +
ε +
ε
3840 n 8448 n
140894208 n
92897280 n
33340423721 19
2826605 17
εn +
ε + O(ε21
+
n ).
210567168
8302787297280 n
Consequently,
lim n9 (αn+1,3 (a) − αn,3 (a)) =
n→∞
127
,
3840
and from this, based on Lemma 1.1, we get
lim n8 (γ(a) − αn,3 (a)) =
n→∞
127
.
30720
(ii) We have
βn,3 (a) − βn+1,3 (a)
ε8n
ε8n
127
127
·
·
−
1
8
30720 (1 − 2 εn )
30720 (1 + 12 εn )8
2555 11 114794663 13 18092183 15
36074257 17
=
ε +
ε +
ε +
ε + O(ε19
n ).
33792 n
704471040 n
92897280 n
210567168 n
= αn,3 (a) − αn+1,3 (a) +
Thus
lim n11 (βn,3 (a) − βn+1,3 (a)) =
n→∞
2555
,
33792
and applying Lemma 1.1, it follows that
lim n10 (βn,3 (a) − γ(a)) =
n→∞
511
.
67584
112
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Alina Sînt m rian
15, 3 (2013)
(iii) We an write that
δn+1,3 (a) − δn,3 (a)
511
511
ε10
ε10
n
n
+
·
·
67584 (1 + 21 εn )10 67584 (1 − 12 εn )10
178161637 13 69794707 15
=
ε +
ε + O(ε17
n ).
704471040 n
92897280 n
= βn+1,3 (a) − βn,3 (a) −
Hene
lim n13 (δn+1,3 (a) − δn,3 (a)) =
n→∞
This, along with Lemma 1.1, gives
178161637
.
704471040
178161637
.
8453652480
lim n12 (γ(a) − δn,3 (a)) =
n→∞
Also, onsidering the sequene in eah of the following parts and using similar arguments as
in Theorem 2.2, we get the indiated limit:
178161637
ηn,3 (a) = δn,3 (a) +
12 , n ∈ N,
8453652480 a + n − 12
lim n14 (ηn,3 (a) − γ(a)) =
n→∞
θn,3 (a) = ηn,3 (a) −
8191
98304 a + n −
1 14
2
, n ∈ N,
lim n16 (γ(a) − θn,3 (a)) =
n→∞
λn,3 (a) = θn,3 (a) +
118518239
267386880 a + n −
1 16
2
lim n18 (λn,3 (a) − γ(a)) =
91282102592903
;
29890034270208
91282102592903
29890034270208 a + n −
1 18
2
lim n20 (γ(a) − µn,3 (a)) =
n→∞
118518239
;
267386880
, n ∈ N,
n→∞
µn,3 (a) = λn,3 (a) −
8191
;
98304
, n ∈ N,
91546277357
.
3460300800
We remark the pattern in forming the sequenes from Theorem 2.2 and those mentioned above.
For example, the general term of the sequene (µn,3 (a))n∈N an be written in the form
µn,3 (a) =
n
X
1
1 B2
1
1
23 − 1 B4
− ·
·
·
·
−
4
3
1 2
a+k−1 2 2
2
4
a+n− 2
a + n − 12
k=1
!
9
X
a + n − 12
ck,3
25 − 1 B6
1
−
·
+
·
− ln
2k ,
a
25
6 a a + n − 1 5
a+n− 1
2
k=4
2
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Euler's onstant, new lasses of sequenes and estimates
15, 3 (2013)
with
ck,3
 2k−1
2
−1




2k−1

2

 22k−1 − 1
=
 22k−1



22k−1 − 1



22k−1
113
B2k
,
if k = 3p + 1, p ∈ N,
2k
B2k
if k = 3p + 2, p ∈ N,
·
,
2k
5
k3
2 − 1 B6
B2k
3
, if k = 3p + 3, p ∈ N,
− 5 ·
·
+
2k
k
2
6
·
where B2k is the 2kth Bernoulli number. Related to this remark, see also [16, Remark 3.4℄, [18, p.
71, Remark 2.1.3; pp. 100, 101, Remark 3.1.6℄.
For Euler's onstant γ = 0.5772156649 . . . we obtain, for example:
α2,3 (1) = 0.5772135222 . . . ;
α3,3 (1) = 0.5772155039 . . . ;
β2,3 (1) = 0.5772162315 . . . ;
β3,3 (1) = 0.5772156875 . . . ;
δ2,3 (1) = 0.5772154387 . . . ;
δ3,3 (1) = 0.5772156601 . . . ;
η2,3 (1) = 0.5772157923 . . . ;
η3,3 (1) = 0.5772156663 . . . ;
θ2,3 (1) = 0.5772155686 . . . ;
θ3,3 (1) = 0.5772156643 . . . ;
λ2,3 (1) = 0.5772157590 . . . ;
λ3,3 (1) = 0.5772156651 . . . ;
µ2,3 (1) = 0.5772155491 . . . ;
µ3,3 (1) = 0.5772156647 . . . .
As an be seen, θ3,3 (1) and µ3,3 (1) are aurate to nine deimal plaes in approximating γ.
Conluding,
the
following remark an be made. Let a ∈ (0, +∞) and q ∈ N \ {1}. Let
2q−1
B2q
1
·
n0 = min n ∈ Na + n − 12 + 2 22q−1−1 · 2q
. We onsider the sequene
2q−1 > 0
(a+n− 12 )
(αn,q (a))n≥n0 dened by
αn,q (a) =
q−1
n
X
X 22k−1 − 1 B2k
1
1
·
−
·
2k
2k−1
a+k−1
2
2k
a + n − 21
k=1
k=1
a+n−
a
− ln
1
2
22q−1 − 1 B2q
1
·
+
·
22q−1
2q a a + n − 1 2q−1
2
!
,
for every n ∈ N, n ≥ n0 . Clearly, lim αn,q (a) = γ(a). Based on the sequene (αn,q (a))n≥n0 , a
n→∞
lass of sequenes onvergent to γ(a) an be onsidered, namely {(αn,q,r (a))n≥n0 |r ∈ N, r ≥ q+1},
where
r
X
ck,q
αn,q,r (a) = αn,q (a) −
,
1 2k
k=q+1 a + n − 2
for every n ∈ N, n ≥ n0 , with
ck,q
 2k−1
2
−1



2k−1
2
=
2k−1

2
−1


22k−1
B2k
if k ∈ {qp + 1, qp + 2, . . . , qp + q − 1}, p ∈ N,
,
2k
2q−1
k
2
− 1 B2q q
B2k
q
, if k = qp + q, p ∈ N.
− 2q−1 ·
·
+
2k
k
2
2q
·
114
CUBO
Alina Sînt m rian
15, 3 (2013)
In this setion we have obtained that the sequene (αn,q (a))n∈N onverges to γ(a) as n−(2q+2)
and that the sequene (αn,q,r (a))n∈N onverges to γ(a) as n−(2r+2) , for q ∈ {2, 3} and r ∈
{q + 1, q + 2, q + 3, q + 4, q + 5, q + 6}.
3
Best bounds
Let (αn )n∈N be the sequene dened by αn = αn,2 (1). In part (i) of Theorem 2.1 we have proved
that
31
.
lim n6 (αn − γ) =
(3)
n→∞
8064
Proposition 3.1.
We have γ < αn+1 < αn , for every n ∈ N.
Proof. We have
αn+1 − αn
4
4
960 n + 32 − 7
960 n + 12 − 7
1
1
1
−
+
− ln
+ ln
.
=
3
3
n + 1 24 n + 3 2 24 n + 1 2
n + 23
n + 21
2
2
Considering the funtion h : [1, +∞) → R, dened by
4
4
960 x + 32 − 7
960 x + 21 − 7
1
1
1
h(x) =
−
+
− ln
+ ln
,
3 3
1 3
x + 1 24 x + 3 2 24 x + 1 2
x
+
x
+
2
2
2
2
and dierentiating it, we obtain that
1
1
1
+
−
3
3 3
(x + 1)2
12 x + 2
12 x + 12
3
3
3840 x + 32
3840 x + 12
3
3
+
−
−
+
3
3 4
1 4
x
+
x
+
960 x + 2 − 7
960 x + 2 − 7
2
h ′ (x) = −
1
2
= (28569600x8 + 228556800x7 + 783330048x6 + 1500185088x5 + 1754428416x4
+1282873344x3 + 573399572x2 + 143606824x + 15502847)
/[3(x + 1)2 (2x + 1)3 (2x + 3)3 (960x4 + 1920x3 + 1440x2 + 480x + 53)
×(960x4 + 5760x3 + 12960x2 + 12960x + 4853)] > 0,
for every x ∈ [1, +∞). It follows that the funtion h is stritly inreasing on [1, +∞). Also, one
an observe that limx→∞ h(x) = 0. These imply that h(x) < 0, for every x ∈ [1, +∞). Therefore
αn+1 − αn < 0, for every n ∈ N, i.e. the sequene (αn )n∈N is stritly dereasing. Beause
limn→∞ αn = γ, we onlude that γ < αn+1 < αn , for every n ∈ N.
Now we give our rst main result of this setion.
CUBO
15, 3 (2013)
Theorem 3.2.
Euler's onstant, new lasses of sequenes and estimates
Let c =
q
31
53
−ln
8064( 54
6
4853
3240 −γ
)
115
. We have
31
31
≤ αn − γ <
6 ,
8064(n + c − 1)6
8064 n + 12
for every n ∈ N. Moreover, the onstants c − 1 and
1
2
are the best possible with this property.
Proof. Note that h is the funtion from the proof of Proposition 3.1. Let (un )n∈N be the sequene
dened by
31
.
un = αn −
8064(n + c − 1)6
We have
un+1 − un = αn+1 − αn −
31
31
+
.
6
8064(n + c)
8064(n + c − 1)6
We onsider the funtion f : [1, +∞) → R dened by
f(x) = h(x) −
31
31
+
.
6
8064(x + c)
8064(x + c − 1)6
Dierentiating, we get that
f ′ (x) = h ′ (x) +
= [(x − 2)
31
31
−
1344(x + c)7
1344(x + c − 1)7
20
X
ak xk + a]/[1344(x + 1)2 (2x + 1)3 (2x + 3)3
k=0
×(960x4 + 1920x3 + 1440x2 + 480x + 53)
×(960x4 + 5760x3 + 12960x2 + 12960x + 4853)(x + c)7 (x + c − 1)7 ].
One an verify that ai > 0, i ∈ {0, 1, . . . , 20} and a > 0. It follows that f ′ (x) > 0, for
every x ∈ [2, +∞). Hene, the funtion f is stritly inreasing on [2, +∞). Also, one an see that
limx→∞ f(x) = 0. From these we obtain that f(x) < 0, for every x ∈ [2, +∞). So, un+1 −un < 0, for
every n ≥ 2, i.e. the sequene (un )n≥2 is stritly dereasing. Having in view that limn→∞ un = γ,
we are able to write that γ < un , for every n ≥ 2. Consequently,
31
≤ αn − γ,
8064(n + c − 1)6
for every n ∈ N, and the onstant c − 1 is the best possible with this property (the equality holds
only when n = 1).
Let (vn )n∈N be the sequene dened by
vn = αn −
31
8064 n +
1 6
2
.
116
CUBO
Alina Sînt m rian
15, 3 (2013)
Then
vn+1 − vn = αn+1 − αn −
31
8064 n +
+
3 6
2
31
8064 n +
1 6
2
.
Dierentiating the funtion g : [1, +∞) → R, dened by
31
g(x) = h(x) −
8064 x +
3 6
2
31
+
8064 x +
1 6
2
,
we obtain that
g ′ (x) = h ′ (x) +
31
1344 x +
3 7
2
31
−
1344 x +
1 7
2
13
= −(93776707584x14 + 1312873906176x
+ 8441879298048x12
+33033108455424x11 + 87842644390912x10 + 167855050098688x9
+237559279782912x8 + 252802412814336x7 + 203105932312256x6
+122459215673472x5 + 54452798252624x4 + 17269355301696x3
+3675508601216x2 + 465873090688x + 26069935939)
/[21(x + 1)2 (2x + 1)7 (2x + 3)7 (960x4 + 1920x3 + 1440x2 + 480x + 53)
×(960x4 + 5760x3 + 12960x2 + 12960x + 4853)].
Thus g ′ (x) < 0, for every x ∈ [1, +∞). Hereby, the funtion g is stritly dereasing on [1, +∞).
Clearly, limx→∞ g(x) = 0. These yield g(x) > 0, for every x ∈ [1, +∞). Then vn+1 − vn > 0, for
every n ∈ N, whih means that the sequene (vn )n∈N is stritly inreasing. Sine limn→∞ vn = γ,
it follows that vn < γ, for every n ∈ N. We an therefore write that
αn − γ <
31
8064 n +
1 6
2
,
(4)
for every n ∈ N. It remains to prove that the onstant 12 is the best possible with the property
that the above inequality (4) holds for every n ∈ N, and this an be ahieved as follows. We have
just proved that
s
6
31
1
−n> ,
8064(αn − γ)
2
n ∈ N.
(5)
CUBO
Euler's onstant, new lasses of sequenes and estimates
15, 3 (2013)
117
Using (1) and (2), we get that
7
1
αn − γ = Hn −
− ln n + 2 −
3
1 2
24 n + 2
960 n + 12
1
!
−γ
!
7
1
= ψ(n + 1) −
2 − ln n + 2 −
3
24 n + 12
960 n + 12
1
1
1
1
1
1
=
+
−
+
+
O
−
2n 12n2
120n4
252n6
240n8
n10
!
1
1
7
−
− ln 1 + 2n −
1 2
1 3
24n2 1 + 2n
960n4 1 + 2n
1
31
31
.
=
−
+
O
6
7
8064n
2688n
n8
1
Let An =
we have
q
6
s
6
31
,
8064n6 (αn −γ)
(6)
n ∈ N. Clearly, lim An = 1, having in view (3). Then, based on (6),
n→∞
!
1
−1
8064 6
31 n (αn − γ)
!
n
31
− n = n(An − 1) = P5
k
8064(αn − γ)
k=0 An
= P5
n
1−
k
k=0 An
= P5
1
k
k=0 An
·
3
n
1
+O
3+O
1−
3
n
1
n
+O
1
n2
1
n2
−1
→
1
1
·3=
6
2
(n → ∞).
(7)
Indeed, from (5) and (7) we obtain that 12 is the best onstant with the property that inequality
(4) holds for every n ∈ N, and now the proof is omplete.
n )n∈N be the sequene dened by α
n = αn,3 (1). In part (i) of Theorem 2.2 we have
Let (α
proved that
127
n) =
.
lim n8 (γ − α
(8)
n→∞
30720
Proposition 3.3.
n+1 < γ, for every n ∈ N.
n < α
We have α
Proof. We have
n+1 − α
n =
α
1
1
7
7
1
−
+
+
−
n + 1 24 n + 3 2 24 n + 1 2 960 n + 3 4 960 n + 1 4
2
2
2
2
3 6
1 6
8064 n + 2 + 31
8064 n + 2 + 31
+ ln
.
− ln
5
3 5
n+ 2
n + 21
118
Alina Sînt m rian
CUBO
15, 3 (2013)
: [1, +∞) → R, dened by
Considering the funtion h
h(x)
=
1
1
7
7
1
−
+
+
−
4
2
2
4
x + 1 24 x + 3
24 x + 21
960 x + 32
960 x + 12
2
6
6
8064 x + 32 + 31
8064 x + 12 + 31
− ln
+ ln
,
5
5
x + 32
x + 12
and dierentiating it, we obtain that
1
1
1
7
7
+
−
−
+
5
3
3
5
2
(x + 1)
12 x + 23
12 x + 21
240 x + 32
240 x + 12
5
5
48384 x + 21
48384 x + 32
5
5
+
−
+
−
6
6
3
1
8064 x + 32 + 31 x + 2
8064 x + 12 + 31 x + 2
′ (x) = −
h
= −(297308454912x14 + 4162318368768x13 + 26769400971264x12
+104792256479232x11 + 278852137150464x10 + 533369371889664x9
+755912773435392x8 + 806006485057536x7 + 649383667564032x6
+393129697342464x5 + 175861357984144x4 + 56282483209792x3
+12150419739472x2 + 1576326994464x + 91873672505)
/[15(x + 1)2 (2x + 1)5 (2x + 3)5
×(8064x6 + 24192x5 + 30240x4 + 20160x3 + 7560x2 + 1512x + 157)
×(8064x6 + 72576x5 + 272160x4 + 544320x3
+612360x2 + 367416x + 91885)] < 0,
is stritly dereasing on [1, +∞). Also, one
for every x ∈ [1, +∞). It follows that the funtion h
an observe that limx→∞ h(x) = 0. These imply that h(x)
> 0, for every x ∈ [1, +∞). Therefore
n > 0, for every n ∈ N, i.e. the sequene (α
n )n∈N is stritly inreasing. Beause
n+1 − α
α
n = γ, we onlude that α
n+1 < γ, for every n ∈ N.
n < α
limn→∞ α
Now we give our seond main result of this setion.
q
127
Theorem 3.4. Let c = 8 30720(γ− 4777
91885 . We have
4860 −ln 61236 )
127
127
n <
≤γ−α
8 ,
8
30720(n + c − 1)
30720 n + 21
for every n ∈ N. Moreover, the onstants c − 1 and
1
2
are the best possible with this property.
is the funtion from the proof of Proposition 3.3. Let (u
n )n∈N be the sequene
Proof. Note that h
dened by
127
n +
n = α
u
.
30720(n + c − 1)8
CUBO
15, 3 (2013)
Euler's onstant, new lasses of sequenes and estimates
119
We have
n+1 − u
n = α
n+1 − α
n +
u
127
127
−
.
8
30720(n + c)
30720(n + c − 1)8
We onsider the funtion f : [1, +∞) → R dened by
= h(x)
f(x)
+
127
127
−
.
30720(x + c)8 30720(x + c − 1)8
Dierentiating, we get that
′ (x) −
f′ (x) = h
127
127
+
9
3840(x + c)
3840(x + c − 1)9
= −[(x − 2)
30
X
k xk + a
]/[3840(x + 1)2 (2x + 1)5 (2x + 3)5
a
k=0
6
×(8064x + 24192x5 + 30240x4 + 20160x3 + 7560x2 + 1512x + 157)
×(8064x6 + 72576x5 + 272160x4 + 544320x3 + 612360x2
+367416x + 91885)(x + c)9 (x + c − 1)9 ].
i > 0, i ∈ {0, 1, . . . , 30} and a
> 0. It follows that f′ (x) < 0, for
One an verify that a
every x ∈ [2, +∞). Hene, the funtion f is stritly dereasing on [2, +∞). Also, one an see that
> 0, for every x ∈ [2, +∞). So, u
= 0. From these we obtain that f(x)
n+1 −u
n > 0, for
limx→∞ f(x)
n )n≥2 is stritly inreasing. Having in view that limn→∞ u
n = γ,
every n ≥ 2, i.e. the sequene (u
n < γ, for every n ≥ 2. Consequently,
we are able to write that u
127
n,
≤γ−α
30720(n + c − 1)8
for every n ∈ N, and the onstant c − 1 is the best possible with this property (the equality holds
only when n = 1).
Let (vn )n∈N be the sequene dened by
vn = α
n +
127
30720 n +
Then
vn+1 − vn = α
n+1 − α
n +
127
30720 n +
: [1, +∞) → R, dened by
Dierentiating the funtion g
(x) = h(x)
g
+
127
30720 x +
3 8
2
−
1 8
2
.
−
3 8
2
127
30720 n +
127
30720 x +
1 8
2
,
1 8
2
.
120
CUBO
Alina Sînt m rian
15, 3 (2013)
we obtain that
′ (x) −
′ (x) = h
g
127
3840 x +
+
3 9
2
20
= (212667885158400x
127
3840 x +
1 9
2
+ 4253357703168000x19 + 40151062789226496x18
+237836352044924928x17 + 991344446628691968x16 + 3090222937974767616x15
+7473501796931665920x14 + 14356208148056506368x13 + 22242021354079121408x12
+28060052688951263232x11 + 28976739296839394304x10 + 24531604126817085440x9
+16993882015446322432x8 + 9580270969643116544x7 + 4353524933287391360x6
+1571495149494432512x5 + 440878222795013392x4 + 93004870693412928x3
+13980891645509980x2 + 1352763769145912x + 64676820697555)
/[15(x + 1)2 (2x + 1)9 (2x + 3)9
×(8064x6 + 24192x5 + 30240x4 + 20160x3 + 7560x2 + 1512x + 157)
×(8064x6 + 72576x5 + 272160x4 + 544320x3 + 612360x2 + 367416x + 91885)].
′ (x) > 0, for every x ∈ [1, +∞). Hereby, the funtion g
is stritly inreasing on [1, +∞).
Thus g
(x) = 0. These yield g
(x) < 0, for every x ∈ [1, +∞). Then vn+1 − Clearly, limx→∞ g
vn < 0, for
vn )n∈N is stritly dereasing. Sine limn→∞ vn = γ,
every n ∈ N, whih means that the sequene (
it follows that γ < vn , for every n ∈ N. We an therefore write that
n <
γ−α
127
30720 n +
1 8
2
(9)
,
for every n ∈ N. It remains to prove that the onstant 12 is the best possible with the property
that the above inequality (9) holds for every n ∈ N, and this an be ahieved as follows. We have
just proved that
s
1
127
8
(10)
− n > , n ∈ N.
n)
30720(γ − α
2
Using (1) and (2), we get that
n
γ−α
1
31
1
= γ − Hn +
−
+ ln n + +
2
4
2 8064 n + 1 5
24 n + 12
960 n + 12
2
7
!
1
31
= −ψ(n + 1) +
−
+ ln n + 2 +
5
1 2
1 4
24 n + 2
960 n + 2
8064 n + 12
1
1
1
1
1
1
1
−
+
−
+
+O
+
= −
2n 12n2
120n4
252n6
240n8
132n10
n12
1
1
7
31
1
+
−
+ ln 1 + 2n +
1 2
1 4
2
4
6
24n 1 + 2n
960n 1 + 2n
8064n 1 +
127
127
1
=
.
−
+O
30720n8
7680n9
n10
7
1 5
2n
!
!
(11)
CUBO
Euler's onstant, new lasses of sequenes and estimates
15, 3 (2013)
n =
Let A
q
8
(11), we have
s
8
127
n) ,
30720n8 (γ−α
121
n = 1, having in view (8). Then, based on
n ∈ N. Clearly, lim A
n→∞
!
1
−1
30720 8
127 n (γ − αn )
!
127
n − 1) = P n
− n = n(A
7
k
n)
30720(γ − α
k=0 An
= P7
n
k
k=0 An
= P7
1
1−
·
4
n
1
+O
4+O
k 1 −
k=0 An
4
n
1
n
+O
1
n2
1
n2
−1
→
1
1
·4 =
8
2
(n → ∞).
(12)
Combining (10) and (12) we obtain that 12 is the best onstant with the property that inequality
(9) holds for every n ∈ N, and now the proof is omplete.
Reeived: September 2012. Aepted: September 2013.
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