arXiv:1002.3856v1 [math.CA] 20 Feb 2010 n[1 p 0–0] yconsidering by 105–106], pp. [11, In ψ h first the am ucinΓ( function gamma e 1 .28 ..] where 6.3.2], 258, p. [1, see scle amncsre.The series. harmonic called is constant. sity. ( x h series The 2000 n[7,tes-aldFae’ nqaiyi ieauewsgvnby given was literature in inequality Franel’s so-called the [17], In e od n phrases. and words Key h uhr eespotdi atb h cec Foundation Science the by part in supported were authors The hspprwstpstusing typeset was paper This eoe h s ucin h oaihi derivative logarithmic the function, psi the denotes ) ahmtc ujc Classification. Subject Mathematics e hr obeieult o onighroi ubr a numbers harmonic bounding for n inequality the double and constant, sharp Euler-Mascheroni new or numbers harmonic ing Abstract. svld iheult ntelf-adsd nywhen only side left-hand the in equality with valid, is 2(7 n ∈ 2 em ftehroi eis a egvnaayial by analytically given be may series, harmonic the of terms − γ − HR ONSFRHROI NUMBERS HARMONIC FOR BOUNDS SHARP N − 12 12 h obeinequality double the , 1 γ n ) 2 and 2(7 + nteppr efis uvysm eut nieulte for inequalities on results some survey first we paper, the In 2 x 1 n 5 6 hc a edfie by defined be may which ) I n r h etpossible. best the are − − 1 = 12 amncnme,pifnto,sapieult,Euler-Ma inequality, sharp function, psi number, harmonic 8 Γ( H n 1 γ Z ) x 2 1 / ( A + 1 EGQ N A-IGUO BAI-NI AND QI FENG γ /n 1 = ) (2 n H < M = ) 0 = γ  n S 1. t amncnumber harmonic -th − 2 1 Z -L x 1 A ( 0 1) X + . i n T Introduction − ∞ 57721566 =1 n E ) ≤ X. 3 1  t − x 1 i rmr 61;Scnay33B15. Secondary 26D15; Primary x 1 H + − ln  = ( 1 1 · · · n e n ) − γ d − − t + + · · · x d ln < γ n ψ ,x> x t, 1 ln = n ( sElrMshrn osatand constant Euler-Mascheroni is + − n 2 1) + · · · 2 1 n 1 n n − n , n − H 0 ,weetescalars the where 1, = , H < γ . ( fTajnPltcncUniver- Polytechnic Tianjin of Γ Γ( n ′ ( ( for ) n x ∈ x − ) ) (5) ) eetbiha establish we n N 12 ftecasclEuler classical the of olw:For follows: s . n n 2 1 ∈ 6 + bound- N / 5 h u of sum the , scheroni (4) (2) (1) (3) 2 F. QI AND B.-N. GUO

1 and 0 < In < 2 , where [t] denotes the largest integer less tan or equal to t, it was established that 1 integer n we have 1 1 summation formula, some general inequalities for the n-th harmonic number H(n) are established, including recovery of the inequality (10). In [25], the problems above-mentioned was solved once again by employing

q−1 ∞ 1 1 B B (x) H(n) = ln n + γ + 2i 2q d x (13) 2n − 2 in2i − Z x2q Xi=1 n and ∞ q B2q−1(x) ( 1) B2q 2q d x< − 2q , (14) Zn x 2qn where n and q are positive integers, Bi(x) are Bernoulli polynomials and B2i = B2i(0) denote Bernoulli numbers for i N. For definitions of Bi(x) and B2i, please refer to [1, p. 804]. ∈ In [23], the inequality (10) was verified again by calculus. SHARPBOUNDSFORHARMONICNUMBERS 3

In [13], by utilizing Euler-Maclaurin summation formula, the following general result was obtained: m 1 1 1 B − 1 H(n) = ln n + γ + + 2(i 1) + O . (15) 2n − 12n2 2 (i 1)n2(i−1)  n2m  Xi=3 − See also [15, p. 77]. In fact, this is equivalent to the formula in [1, p. 259, 6.3.18]. In [22], by considering the decreasing monotonicity of the sequence 1 xn = ∞ 2n, (16) ( 1)k−1 1 − k=n+1 − k it was shown that the best constants P a and b such that ∞ 1 k−1 1 1 ( 1) < (17) 2n + a ≤ − k 2n + b k=Xn+1 for n 1 are a = 1 2 and b = 1. ≥ 1−ln 2 − In [4, Theorem 2.8] and [19], alternative sharp bounds for H(n) were presented: For n N, ∈ 1 + ln √e 1 ln e1/(n+1) 1 H(n) <γ ln e1/(n+1) 1 . (18) − − − ≤ − −


The constants 1 + ln √e 1 and γ in (18) are the best possible. This improves − the result in [3, pp. 386–387].
In [20], it was established that

1 − ln n + + γ

2. Proof of Theorem 1 We now are in a position to prove Theorem 1. Let 1 f(x)= 12x2 (22) ln x +1/2x ψ(x + 1) − − for x (0, ). An easy computation gives ∈ ∞ 2 ′ 2 ′ 4x ψ (x + 1) 4x +2 4x g(x) f (x)= − 24x = , [2x ln x 2xψ(x + 1) + 1]2 − [2x ln x 2xψ(x + 1) + 1]2 − − where 2 ′ 1 1 1 g(x)= ψ (x + 1) + 24x ψ(x + 1) ln x . (23) − x 2x2 −  − − 2x In [2, Theorem 8], the functions 2n 1 1 B F (x) = ln Γ(x) x ln x + x ln(2π) 2j (24) n −  − 2 − 2 − 2j(2j 1)x2j−1 Xj=1 − and 2n+1 1 1 B G (x)= ln Γ(x)+ x ln x x + ln(2π)+ 2j (25) n −  − 2 − 2 2j(2j 1)x2j−1 Xj=1 − for n 0 were proved to be completely monotonic on (0, ). This generalizes [16, Theorem≥ 1] which states that the functions F (x) and G ∞(x) are convex on (0, ). n n ∞ The complete monotonicity of Fn(x) and Gn(x) was proved in [12, Theorem 2] once again. In particular, the functions 1 1 F (x) = ln Γ(x) x ln x + x ln(2π) 2 −  − 2 − 2 (26) 1 1 1 1 + + − 12x 360x3 − 1260x5 1680x7 and 1 1 G (x)= ln Γ(x)+ x ln x x + ln(2π) 1 −  − 2 − 2 (27) 1 1 1 + + 12x − 360x3 1260x5 are completely monotonic on (0, ). Therefore, we have ∞ 1260x5 + 210x4 21x2 + 10 ln x − < ψ(x) − 2520x6 2520x7 + 420x6 42x4 + 20x2 21 < ln x − − (28) − 5040x8 and 8 7 6 4 2 210x + 105x + 35x 7x +5x 7 ′ − − < ψ (x) 210x9 210x6 + 105x5 + 35x4 7x2 +5 < − (29) 210x7 on (0, ). From this, it follows that ∞ SHARPBOUNDSFORHARMONICNUMBERS 5

1 1 ln x + ψ(x + 1) = ln x ψ(x) 2x − − 2x − 1260x5 + 210x4 21x2 + 10 1 10 21x2 + 210x4 < − = − (30) 2520x6 − 2x 2520x6 and

2 4 2 ′ 1 1 1 10 21x + 210x g(x) > ψ (x) + − − x2 − x 2x2 − 264600x11
210x8 + 105x7 + 35x6 7x4 +5x2 7 1 1 > − − 210x9 − x2 − x 2 1 10 21x2 + 210x4 + − 2x2 − 264600x11
1659x4 8400x2 100 = − − 264600x11 1659(x 3)4 + 19908(x 3)3 + 81186(x 3)2 + 128772(x 3) + 58679 = − − − − . 264600x11 Hence, the function g(x) is positive on [3, ). So the derivative f ′(x) > 0 on [3, ), that is, the function f(x) is strictly increasing∞ on [3, ). ∞ It is easy to obtain ∞ 2(7 12γ) f(1) = − =0.9507 , 2γ 1 ··· − 4(48γ + 48 ln 2 61) f(2) = − =1.1090 , 5 4γ 4 ln 2 ··· − − 3(108γ + 108 ln 3 181) f(3) = − =1.1549 . 5 3γ 3 ln 3 ··· − − This means that the sequence f(n) for n N is strictly increasing. Employing the inequality (30) yields ∈

2520x6 12x2 21x2 10 6 f(x) > 12x2 = − 10 21x2 + 210x4 − 10 21 x2 + 210x
4 → 5 − − as x . Utilizing the right-hand side inequality in (28) leads to → ∞ 1 f(x)= 12x2 ln x 1/2x ψ(x) − − − 1 < 12x2 (2520x7 + 420x6 42x4 + 20x2 21)/5040x8 1/2x − − − − 12x2 42x4 20x2 + 21 = − 420x6 42x4 + 20x2 21
6 − − → 5 6 as x . As a result, it follows that lim →∞ f(x) = . Therefore, it is derived → ∞ x 5 that f(1) f(n) < 6 for n N, equivalently, ≤ 5 ∈ 2(7 12γ) 1 6 − 12n2 < 2γ 1 ≤ ln n +1/2n ψ(n + 1) − 5 − − 6 F. QI AND B.-N. GUO which can be rearranged as 1 1 1 ln n + ψ(n + 1) > . 12n2 + 2(7 12γ)/(2γ 1) ≥ 2n − 12n2 +6/5 − − Combining this with (2) yields (21). The proof of Theorem 1 is proved.

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(F. Qi) Department of Mathematics, College of Science, Tianjin Polytechnic Univer- sity, Tianjin City, 300160, China E-mail address: [email protected], [email protected], [email protected] URL: http://qifeng618.spaces.live.com

(B.-N. Guo) School of Mathematics and Informatics, Henan Polytechnic University, Jiaozuo City, Henan Province, 454010, China E-mail address: [email protected], [email protected]