Note of the constants of Landau
1Emil C. Popa
Abstract
We establish an improvement for an inequality of Dejun Zhao for the Landau constants.
2000 Mathematics Subject Classification: 26D20, 33B15.
Key words and phrases: Landau’s constants, Riemann-Zeta function.
1 Introduction
The Landau’s constants are defined by
Gn=
n
X
k=0
1 16k
2k
k
2
, n≥0 and play an important role in complex analysis.
1Received 2 December, 2009
Accepted for publication (in revised form) 6 January, 2010
113
D. Zhao [5] proved the following several sharp inequalities for Gn : 1
π ln(n+ 1) +c0− 1
4π(n+ 1) + 5
192π(n+ 1)2 < Gn (1)
≤ 1
π ln(n+ 1) +c0− 1
4π(n+ 1) + 5
192π(n+ 1)2 + 3
128π(n+ 1)3, where c0 = 1
π(γ+ ln 16), and obtained a new asymptotic formula for Gn Gn=1
π ln(n+1)+c0− 1
4π(n+1)+ 5
192π(n+1)2+O
1 (n+1)3
(2) .
In this paper we will establish an improvement for the left-hand inequality of Zhao.
Theorem 1 We have for all integer n≥1 (3)
1
πln(n+1)+c0− 1
4π(n+1)+ 5
192π(n+1)2+ 17
256π ζ(4)−
n+1
X
k=1
1 k4
!
< Gn,
where ζ(s) =
∞
X
k=1
k−s is the Riemann-Zeta function.
2 The proof of the Theorem 1
Using [4] and [5] we have
πp2n< 8n+ 3 8n2+ 5n+ 1, where pn = (2n)!
4n(n!)2. Hence
π(Gn−Gn−1)< 8n+ 3 8n2+ 5n+ 1.
Let now xn=Gn−1
π ln(n+1)−c0+ 1
4π(n+ 1)− 5
192π(n+1)2−A
π ζ(4)−
n+1
X
k=1
1 k4
! ,
where A is an undetermined constant.
We get for n≥1
π(xn−xn−1) = πp2n−ln
1 + 1 n
+1
4 1
n+ 1 − 1 n
− 5 192π
1
(n+ 1)2 − 1 n2
+ A
(n+ 1)4 or
π(xn−xn−1) < 8n+ 3
8n2+ 5n+ 1 −ln
1 + 1 n
+1
4 1
n+ 1 − 1 n
− 5 192π
1
(n+ 1)2 − 1 n2
+ A
(n+ 1)4. Next we consider the function f(x) defined by
f(x) = 8x+ 3
8x2+ 5x+ 1 −ln
1 + 1 x
+1
4 1
x+ 1 − 1 x
− 5 192π
1
(x+ 1)2 − 1 x2
+ A
(x+ 1)4 and we have
f0(x) = Q(x)
4x3(x+ 1)3(8x2+ 5x+ 1)2 − 4A (x+ 1)5, where
Q(x) = 68x5+ 2545
24 x4+1067
24 x3+ 19
12x2− 41 24x− 5
24. Now we denote
f0(x) = H(x)
4x3(x+ 1)5(8x2+ 5x+ 1)2
where
H(x) = (68−1024A)x7+
5809
24 −1280A
x6
+
7789
24 −656A
x5+
4717
24 −160A
x4
+
1102
24 −16A
x3− 49
24x2− 51 24x− 5
24. For A= 17
256 we obtain H(x) = 3769
24 x6+13487
48 x5+ 2231 12 x4 + 2153
48 x3− 49
29x2− 51 24x− 5
24 >0, for x≥1.
Hence f0(x) > 0 for x ≥ 0 and lim
x→∞f(x) = 0. It implies f(x) < 0 for x ≥1. Using the Watson asymptotic formula (see [3]) we have xn →0. In conclusion xn →0,xn>0 forn → ∞and the inequality of the Theorem 1 holds.
References
[1] H. Alzer, Inequalities for the constants of Landau and Lebesque, J. Com- put. Appl. Math., 139, Issue 2, 2002, 215-230.
[2] L.P. Falaleev, Inequalities for the Landau constants, Siberian Math. J., 32, 1991, 896-897.
[3] G.N. Watson, The constants of Landau and Lebesque, Quart. J. Math.
Oxford, Ser. 2 (1), 1930, 310-318.
[4] G.M. Zhang,The upper bounds and lower bounds on Wallis’s inequality, Math. Practice Theory, 37(5), 2007, 111-116 (in China).
[5] D. Zhao,Some sharp estimates of the constants of Landau and Lebesque, J. Math. Anal. Appl. 349, 2009, 68-73.
Emil C. Popa
”Lucian Blaga” University Department of Mathematics Sibiu, Romania
e-mail: [email protected] [email protected]
