Scientia Magna Vol. 1 (2005), No. 2, 91-95
Inequalities for the polygamma functions with application1 Chaoping Chen Department of Applied Mathmatics, Hennan Polytechnic University Jiaozuo, Hennan, P. R. China Abstract We present some inequalities for the polygamma funtions. As an application, we give n P 1 the upper and lower bounds for the expression − ln n − γ, where γ = 0.57721 · · · is the Euler’s k constant. Keywords
k=1
Inequality; Polygamma function; Harmonic sequence; Euler’s constant.
§1. Inequalities for the Polygamma Function The gamma function is usually defined for Rez > 0 by Z ∞ Γ(z) = tz−1 e−t dt. 0
The psi or digamma function, the logarithmic derivative of the gamma function and the polygamma functions can be expressed as ¶ 0 ∞ µ X 1 Γ (z) 1 = −γ + − , ψ(z) = Γ(z) 1+k z+k k=0
ψ n (z) = (−1)n+1 n!
∞ X k=0
1 (z + k)n+1
for Rez > 0 and n = 1, 2, · · · , where γ = 0.57721 · · · is the Euler’s constant. M. Merkle [2] established the inequality 2N ∞ 2N +1 X X X 1 1 B2k 1 1 B2k + 2+ < < + x 2x x2k+1 (x + k)2 x x2k+1 k=1
k=0
k=1
for all real x > 0 and all integers N ≥ 1, where Bk denotes Bernoulli numbers, defined by ∞
X Bj t = tj . t e − 1 j=0 j! The first five Bernoulli numbers with even indices are B2 = 1 This
1 1 1 1 5 , B4 = − , B6 = , B8 = − , B10 = . 6 30 42 30 66
work is supported in part by SF of Henan Innovation Talents at University of P. R. China
92
Chaoping Chen
No. 2
The following theorem 1 establishes a more general result. Theorem 1. Let m ≥ 0 and n ≥ 1 be integers, then we have for x > 0, 2m+1 2m X B2j 1 X 1 B2j 1 1 − < ψ(x) < ln x − − ln x − 2j 2x 2j x 2x 2j x2j j=1 j=1
and
(1)
2m
X B2j Γ(n + 2j) (n − 1)! n! + n+1 + n x 2x (2j)! xn+2j j=1 < (−1)n+1 ψ (n) (x) <
2m+1 X B2j Γ(n + 2j) (n − 1)! n! + + . xn 2xn+1 (2j)! xn+2j j=1
(2)
Proof. From Binet’s formula [6, p. 103] µ ¶ ¶ Z ∞µ √ 1 t t e−xt ln Γ(x) = x − ln x − x + ln 2π + − 1 + dt, 2 et − 1 2 t2 0 we conclude that
1 ψ(x) = ln x − − 2x
Z 0
∞
µ
t t −1+ t e −1 2
and therefore n+1
(−1)
ψ
(n)
(n − 1)! n! (n) = + n+1 + xn 2x
Z
∞
µ
0
¶
e−xt dt t
t t −1+ et − 1 2
(3)
¶ tn−1 e−xt dt.
(4)
It follows from Problem 154 in Part I, Chapter 4, of [3] that 2m+1 2m X B2j X B2j 2j t t t < t −1+ < t2j (2j)! e − 1 2 (2j)! j=1 j=1
(5)
for all integers m > 0. The inequality (5) can be also found in [4]. From (3) and (5) we conclude (1), and we obtain (2) from (4) and (5). This completes the proof of the theorem 1. Note that ψ(x + 1) = ψ(x) + x1 (see [1, p. 258]), (1) can be written as 2m+1 2m X B2j 1 X 1 1 B2j 1 − < ψ(x + 1) − ln x < − 2j 2x 2j x 2x 2j x2j j=1 j=1
(6)
and (2) can be written as 2m
X B2j Γ(n + 2j) n! (n − 1)! − n+1 + n x 2x (2j)! xn+2j j=1 n+1
< (−1)
ψ
(n)
2m+1 X B2j Γ(n + 2j) (n − 1)! n! (x) < − + . xn 2xn+1 (2j)! xn+2j j=1
(7)
In particular, taking in (6) m = 0 we obtain for x > 0, 1 1 1 < ψ(x + 1) − ln x < − 2 2x 12x 2x
(8)
Vol. 1
93
Inequalities for the polygamma functions with application
and taking in (7) m = 1 and n = 1, we obtain for x > 0 0 1 1 1 1 1 1 1 1 − 3+ − < − ψ (x + 1) < 2 − 3 + 2 5 7 2x 6x 30x 42x x 2x 6x 30x5
(9)
The inequalities (8) and (9) play an important role in the proof of the theorem 2 in Section 2.
§2. Inequalities for Euler’s Constant Euler’s constant γ = 0.57721 · · · is defined by µ ¶ 1 1 1 γ = lim 1 + + + · · · + − ln n . n→∞ 2 3 n It is of interest to investigate the bounds for the expression
n P k=1
1 k
− ln n − γ. The inequality
n
X1 1 1 1 − 2 < − ln n − γ < 2n 8n k 2n k=1
is called in literature Franel’s inequality [3, Ex. 18]. n−1 P 1 It is given in [1, p. 258] that ψ(n) = k − γ, and then we have get k=1
n X 1 − ln n − γ = ψ(n + 1) − ln n. k
(10)
k=1
Taking in (6) x = n we obtain that 2m+1 n 2m X B2j 1 X X 1 1 1 B2j 1 − < − ln n − γ < − . 2j 2n 2j n k 2n 2j n2j j=1 j=1
(11)
k=1
The inequality (11) provides closer bounds for
n P k=1
1 k
− ln n − γ.
L.T´oth [5, p. 264] proposed the following problems: (i) Prove that for every positive integer n we have 1 2n +
2 5
<
n X 1 1 . − ln n − γ < k 2n + 31 k=1
(ii) Show that 25 can be replaced by a slightly smaller number, but that replaced by a slightly larger number. The following Theorem 2 answers the problem due to L.T´oth. Theorem 2. For every positive integer n,
1 3
can not be
n
X1 1 1 < − ln n − γ < , 2n + a i 2n + b i=1 with the best possible constants
(12)
94
Chaoping Chen
a=
1 −2 1−γ
No. 2
and
b=
1 3
Proof. By (10), the inequality (12) can be rearranged as 1 − 2n ≤ a. ψ(n + 1) − ln n
b< Define for x > 0
φ(x) =
1 − 2x. ψ(x + 1) − ln x
Differentiating φ and utilizing (8) and (9) reveals that for x >
12 5
0 1 − ψ (x + 1) − 2(ψ(x + 1) − ln x)2 x µ ¶2 1 1 1 1 1 12 − 5x < 2− 3+ − − 2 = < 0, 2x 6x 30x5 2x 12x2 360x5 0
(ψ(x + 1) − ln x)2 φ (x) =
and then the function φ strictly decreases with x > Straightforward calculation produces φ(1) =
φ(2) = φ(3) =
12 5 .
1 − 2 = 0.36527211862544155 · · · , 1−γ
3 2
1 − 4 = 0.35469600731465752 · · · , − γ − ln 2
11 6
1 − 6 = 0.34898948531361115 · · · . − γ − ln 3
Therefore, the sequence φ(n) =
1 − 2n, ψ(n + 1) − ln n
n∈N
is strictly decreasing. This leads to lim φ(n) < φ(n) ≤ φ(1) =
n→∞
1 − 2. 1−γ
Making use of asymptotic formula of ψ (see [1, p. 259]) ψ(x) = ln x −
1 1 + O(x−4 ) − 2x 12x2
we conclude that lim φ(n) = lim φ(x) = lim
n→∞
x→∞
This completes the proof of the theorem 2.
x→∞
1 3
(x → ∞),
+ O(x−2 ) 1 = . 1 + O(x−1 ) 3
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Inequalities for the polygamma functions with application
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References [1] M. Abramowitz and I.Stegun (Eds), Handbook of Mathematical Function with Formulas, Graphs, and Mathematical Tables, 4th printing, with corrections, Applied Mathematics Series 55, National Bureau of Standards, Washington, 1965. [2] M. Merkle, Logarithmic convexity inequalities for the gemma function, J. Math. Anal. Appl, 203(1996), 369–380. [3] G. P´olya and G. Szeg¨o, Problems and Theorems in analysis, Vol.I and II, SpringerVerlag, Berlin, Heidelberg, 1972. [4] Z. Sasv´ari, Inequalities for binomial coefficients, J. Math. Anal. Appl. 236(1999), 223-226. [5] L. T´oth, E 3432, Amer. Math. Monthly 98(1991), 264; 99(1992), 684-685. [6] Zh-x.Wang and D.-R.Guo, Introduction to Special Function, the Series of Advanced Physics of Peking University, Peking University Press, Beijing, China, 2000(Chinese).