Ramauja J. 44 017, 81-99 The log-behavior of p ad p/ William Y.C. Che 1 ad Ke Y. Zheg 1 Ceter for Applied Mathematics Tiaji Uiversity Tiaji 0007, P. R. Chia Ceter for Combiatorics, LPMC Nakai Uivercity Tiaji 00071, P. R. Chia Email: 1 cheyc@tju.edu.c, kezheg@aliyu.com Abstract Let p deote the partitio fuctio ad let be the differece operator respect to. I this paper, we obtai a lower boud for log 1 p 1/ 1, leadig to a proof of the cojecture of Su o the log-covexity of { p/} 60. Usig the same argumet, it ca be show that for ay real umber α, there exists a iteger α such that the sequece { p/ α } α is log-covex. Moreover, we show that lim 5 log p = π/ 4. Fially, by fidig a upper + boud of log 1 p 1, we establish a iequality o the ratio 1 p 1 p. Keywords: Partitio fuctio, Log-covex sequece, Hardy-Ramauja-Rademacher formula, Lehmer s error boud AMS Subject Classificatios: 05A0 1 Itroductio The objective of this paper is to study the log-behavior of the sequeces p ad p/, where p deotes the umber of partitios of a oegative iteger. A positive sequece {a } 0 is log-covex if it satisfies that for 1, a a 1 a +1 0, 1
ad it is called log-cocave if for 1, a a 1 a +1 0. Let r = p/ ad let be the differece operator respect to. Su [11] cojectured that the sequece {r} 60 is log-covex. Desalvo ad Pak [5] oticed that the log-covexity of {r} 60 ca be derived from a estimate for log r 1, see [5, Fial Remark 7.7]. They also remarked that their approach to boudig log p 1 does ot seem to apply to log r 1. I this paper, we obtai a lower boud for log r 1, leadig to a proof of the log-covexity of {r} 60. Theorem 1.1 The sequece {r} 60 is log-covex. The log-covexity of {r} 60 implies the log-covexity of { p} 6, because the sequece { } 4 is log-covex [11]. It is kow that p = 1. For a lim + combiatorial proof of this fact, see Adrews [1]. Su [11] proposed the cojecture that { p} 6 is strictly decreasig, which has bee proved by Wag ad Zhu [1]. The log-covexity of { p} 6 was also cojectured by Su [11]. It is easy to see that the log-covexity of { p} 6 implies the decreasig property. It should be oted that there is a alterative way to prove the log-covexity of { p} 6. Che, Guo ad Wag [] itroduced the otio of a ratio log-covex sequece ad showed that the ratio log-covexity implies the log-covexity uder a certai iitial coditio. A sequece {a } k is called ratio log-covex if {a +1 /a } k is log-covex, or, equivaletly, for k + 1, log a + log a +1 + log a log a 1 0. Che et al. [4] showed that that for ay r 1, oe ca determie a umber r such that for > r, 1 r 1 r log p is positive. For r =, it ca be show that for 116, log p 1 > 0. Sice log p 1 = log p + log p + 1 + log p log p 1, we see {p} 115 is ratio log-covex. So we are led to the followig assertio. Theorem 1. The sequece { p} 6 is log-covex. Moreover, as poited out by the referee, we may cosider the log-behavior of p/ α for ay real umber α. To this ed, we obtai the followig geeralizatio of Theorems 1.1 ad 1..
Theorem 1. Let α be a real umber. There exists a positive iteger α such that the sequece { p/ α } α is log-covex. We also establish the followig iequality o the ratio 1 p 1 p. Theorem 1.4 For, we have p +1 p + 1 1 + π > 4 1 p 1. 1.1 p Desalvo ad Pak [5] have show that the limit of log p is π/ 4. boudig log p, we derive the followig limit of 5 log p: By lim 5 log p = π/ 4. 1. + From the above relatio 1., it ca be see that the coefficet π 4 i 1.1 is the best possible. The Log-covexity of r I this sectio, we obtai a lower boud of log r 1 ad prove the log-covexity of {r} 60. First, we follow the approach of Desalvo ad Pak to give a expressio of log r 1 as a sum of B 1 ad Ẽ 1, where B 1 makes a major cotributio to log r 1 with Ẽ 1 beig the error term, that is, B 1 coverges to log r 1. The expressios for B ad E will be give later. I this settig, we derive a lower boud of B 1. By Lehmer s error boud, we give a upper boud for Ẽ 1. Combiig the lower boud for B 1 ad the upper boud for Ẽ 1, we are led to a lower boud for log r 1. By provig the positivity of this lower boud for log r 1, we reach the log-covexity of {r} 60. The strict log-covexity of {r} 60 ca be restated as the followig relatio for 61 log r + 1 + log r 1 log r > 0, that is, for 61, log r 1 > 0. For 1 ad ay positive iteger N, the Hardy-Ramauja-Rademacher formula see [, 6, 7, 10] reads p = d N [ A µ k 1 k e µk + 1 + k ] e µ k + R, N,.1 µ µ k=1
where d = π 6, µ = π 6 4 1, A k = k 1 A k, A k is a sum of 4th roots of uity with iitial values A 1 = 1 ad A = 1, R, N is the remaider. Lehmer s error boud see [8, 9] for R, N is give by [ N R, N < π N / sih µ µ N + 1 6 ] N.. µ Let us give a outlie of Desalvo ad Pak s approach to provig the log-cocavity of {p} >5. Settig N = i.1, they expressed p as where T = R = d [ µ d µ They have show that ad 1 1 e µ + 1 µ [ 1 + 1 µ p = T + R,. e µ ],.4 ] e µ + R,. e µ 1 µ + 1 1 + µ log p 1 log T 1 = log log T 1 log It follows that log they use log d µ 1 d µ 1 log-cocavity of {p} >5. 1 1 d 1 µ 1 1 1 µ 1 µ 1 1 + 1 µ 1.5 R 1 < e π 10.6 T 1 e µ 1 < e π 10..7 e µ 1 coverges to log p 1. Fially, e µ 1 to estimate log p 1, leadig to the We shall use a alterative decompositio of p. Settig N = i.1, we ca express p as p = T + R,.8 where R = T = d 1 1 e µ,.9 µ µ d [ 1 + 1 e µ + 1 1 µ µ µ e µ + 1 1 + ] e µ + R,..10 µ 4
Based o the decompositio.8 for p, oe ca express log r 1 as follows: log r 1 = B 1 + Ẽ 1,.11 where B = 1 log T 1 log,.1 ỹ = R/ T,.1 Ẽ = 1 log1 + ỹ..14 The followig lemma will be used to derive a lower boud ad a upper boud of B 1. Lemma.1 Suppose fx has a cotiuous secod derivative for x [ 1, + 1]. The there exists c 1, + 1 such that f 1 = f + 1 + f 1 f = c..15 If fx has a icreasig secod derivative, the 1 < f 1 < + 1..16 Coversely, if fx has a decreasig secod derivative, the + 1 < f 1 < 1..17 Proof. Set ϕx = fx + 1 fx. By the mea value theorem, there exists a umber ξ 1, such that f + 1 + f 1 f = ϕ ϕ 1 = ϕ ξ. Agai, applyig the mea value theorem to ϕ ξ, there exists a umber θ 0, 1 such that ϕ ξ = f ξ + 1 f ξ = ξ + θ. Let c = ξ + θ. The we get.15, which yields.16 ad.17. I order to fid a lower boud for log r 1 ad obtai the limit of 5 log p, we eed the followig lower ad upper bouds for 1 1 log T 1. 5
Lemma. Let 4 logµ 1 B 1 =,.18 + 14 + / 1 4 logµ + 1 5 B = + 14 5 / + 1 1..19 For 40, we have B 1 < 1 1 log T 1 < B..0 Proof. By the defiitio.9, we may write where Thus Sice 1 = log T = f 1 = µ, f = f = f 4 = log d. 1 1 log T 1 = 4 f i, i=1 log µ, logµ 1, 4 f i 1..1 i=1 π 16 4 1 / + 864 4 1 + 6 1, we see that for 1, 1 < 0. Similarly, it ca be checked that for 4, > 0, < 0, ad 4 > 0. Cosequetly, for 4, 1 ad are decreasig, whereas ad 4 are icreasig. Usig Lemma.1, for each i, we ca get a lower boud ad a upper boud for f i 1 i terms of i 1 ad i + 1. For example, 1 + 1 < f 1 1 < 1 1. So, by.1 we fid that 1 1 log T 1 > 1 + 1 + 1 + + 1 + 4 1,. 6
ad where 1 1 log T 1 < 1 1 + + 1 + 1 + 4 + 1,. 1 = 4 1 / 1π 4 1 / + π,.4 4 1 / 6 log µ 7 = + 4 1 + 864 4 1,.5 4π logµ 1 = + µ 1 4 1 4π µ 1 4 1 4π µ 14 1 /,.6 4 = log d..7 Accordig to.4, oe ca check that for, 1 + 1 > A easy computatio shows that for, Substitutig.9 ito.6 yields that + 1 > + 14 + 1π..8 / + 1 4 + / µ 1 > µ..9 logµ + 1 1 540 + 1 4 5 1 6 4 5 1..0 Usig.5 ad.0, we fid that 1 + + 1 > logµ + 1 1 6 logµ 1 + 1 1 + 4 14 5 + 6 1 4 5.1 Apparetly, for, + 1 1 > 1 1, 4 7
so that logµ + 1 1 6 logµ 1 + 1 1 > Sice, for, logµ + 1 1 logµ + 1 1 4 logµ 1 + 1 1 1 1 logµ + 1 1 4 logµ 1 >.. 1 4 1 4 14 5 + 6 1 4 5 > 1,. utilizig.1 ad. yields, for, 4 logµ 1 1 logµ + 1 1 1 + + 1 > +..4 1 1 1 4 Usig.7,.8 ad.4, we deduce that 1 + 1 + 1 + + 1 + 4 1 B 1 > 1 + log d 1π 1 logµ + 1 1..5 1 + 1 4 + / 1 4 Let C be the right had side of.5. By., to prove B 1 < 1 1 log T 1, it is eough to show that C > 0 whe 40. Sice log x < x for x > 0, for µ + 1 1 < π 4 4 4,.6 we get 1 logµ + 1 1 1µ + 1 1 > > 4π..7 1 4 1 4 1 7/ Note that for, 1π 4π + 1 4 + >..8 / 48 1 7/ Combiig.7 ad.8, we see that for, 1 + log d + 1/48 4π C >..9 1 1 7/ It is straightforward to show that the right had side of.9 is positive if 490. For 40 489, it is routie to check that C > 0, ad so C > 0 for 40. It follows from.5 that for 40, 1 1 log T 1 > B 1. 8
To derive the upper boud for 1 1 log T 1, we obtai the followig upper bouds which ca be verified directly. The proofs are omitted. For, 1 1 < 1[4 5], / 6 log µ + 1 9 + 1 < + + 1 1, 4π logµ 1 1 < + µ 1 4 5 1 1 + 1 + 1 < 4π µ 1 4 5 1 4π µ 14 5 / 1, 1 logµ + 1 4 logµ + 1 +, 1 1 4 + 1 4 + 1 < 0. Combiig the above upper bouds, we coclude that for 40, This completes the proof. 1 1 + + 1 + 1 + 4 + 1 < B. The followig lemma gives a upper boud for Ẽ 1. Lemma. For 40, Ẽ 1 < 5 π 4 5 1 e 18..40 Proof. By.14, we fid that for, where Ẽ 1 = 1 1 log1 + ỹ 1 + 1 + 1 log1 + ỹ +1 log1 + ỹ,.41 ỹ = R/ T. To boud Ẽ 1, it is ecessary to boud ỹ. For this purpose, we first cosider R, as defied by.10. Sice d < 1 ad µ >, for 1 we have [ d 1+ 1 e µ + 1 1 µ µ µ < 1 1+e µ µ +1. 9 e µ + 1 1+ ] e µ µ
For N = ad 1, Lehmer s boud. reduces to R, < 4 1 + 4 µ e µ. By the defiitio of R, R < 1 1 + e µ µ + 1 + 4 1 + 4 µ e µ < 5 + 9 µ e µ..4 Recallig the defiitio.9 of T, it follows from.4 that for 1, ỹ < Observe that for, ad µ 5µ e µ + 9e µ 6 e µ..4 dµ 1 5µ e µ + 9e µ 6 < 0,.44 dµ 1 µ > 0..45 Sice 5µ 40e µ40 + 9e µ40 6 < usig.44 ad.45, we deduce that for 40, 5µ e µ + 9e µ 6 < Now, it is clear from.4 ad.46 that for 40, dµ40 1, µ40 dµ 1..46 µ ỹ < e µ..47 I view of.47, for 40, ỹ < e µ40 < 1 5..48 It is kow that log1 + x < x for 0 < x < 1 ad log1 + x < x/1 + x for 1 < x < 0. Thus, for x < 1, log1 + x x 1 x,.49 10
see also [5], ad so it follows from.48 ad.49 that for 40, log1 + ỹ ỹ 1 ỹ 5 4 ỹ..50 Because of.41, we see that for, Ẽ 1 1 1 log1+ỹ 1 + 1 +1 log1+ỹ +1 + log1+ỹ..51 Applyig.50 to.51, we obtai that for 40, Ẽ 1 5 4 Ẽ 1 < 5 4 ỹ 1 1 + ỹ +1 + 1 + ỹ Pluggig.47 ito.5, we ifer that for 40, e µ 1 1 µ+1 + e + 1..5 µ + e..5 But 1 µ e is decreasig for 1. It follows from.5 that for 40, This proves.40. Ẽ 1 < 5 µ 1 1 e. With the aid of Lemma. ad., we are ready to prove the log-covexity of {r} 60. Proof of Theorem 1.1. To prove the strict log-covexity of {r} 60, we proceed to show that for 61, log r 1 > 0. Evidetly, for 40, By Lemma.1, that is, It follows from.1 that log 1 1 log 1 1 log > 0. > log 1, 1 log 1 > + 1 1..54 B 1 = 1 1 log T log 1 1. 1 11
Applyig Lemma. ad.54 to the above relatio, we deduce that for 40, that is, B 1 > B 1 > B1 log 1 + 1 1, 4 log[µ 1] log 1 + + 14 + / 1 1 1..55 By.11 ad Lemma., we fid that for 40, log r 1 > 5 π B 4 5 1 1 e 18..56 It follows from.55 ad.56 that for 40, log r 1 > 4 log[µ 1] log 1 + + 14 + / 1 1 1 5 π 4 5 1 e 18. Let D deote the right had side of the above relatio. Clearly, for 5505, + 14 + > π > 1..57 / 4 + 1 1 To prove that D > 0 for 5505, we wish to show that for 5505, 4 log[µ 1] log 1 + 1 1 1 5 π 4 5 1 1 e 18 >..58 1 Usig the fact that for x > 5504, log x < x 1/4, we deduce that for 5505, ad 4 log[µ 1] < 4 4 µ 1 < 4 4 1 1 log 1 1 < π 11/4 1 < Sice e x > x 6 /70 for x > 0, we see that for, 4 4 4 < 1 6,.59 1 /8..60 1 11/4 1 π 4 5 1 e 18 < 1 π 1 e 18 < 094 1 < 094 1..61 4 Combiig.59,.60 ad.61, we fid that for 5505, 4 log[µ 1] log 1 + 1 1 1 5 π 4 5 1 e 18 1
6 > 1 /8 1 + 11/4 1 10470 1 4 6 > 1 /8 1 11/4 1 > 1. This proves the iequality.58. By.58 ad.57, we obtai that D > 0 for 5505. Verifyig that log r 1 > 0 for 61 5504 completes the proof. Clearly, Theorem 1. is a geeralizatio as well as a uificatio of Theorem 1.1 ad 1.. I fact, it ca be proved i the same maer as the proof of Theorem 1.1. 1 Proof of Theorem 1.. Let α be a real umber. Whe α 0, it is clear that α log-covex. It follows from Theorem 1. that p/ α is log-covex for 6. We ow cosider the case α > 0. A similar argumet to the proof of Theorem 1.1 shows that for 40, log 1 p 1/ 1 α = 1 1 log T + 1 > 4 log[µ 1] + 14 + / 1 1 log1 + y log 1 1 α 1 α log 1 1 + α 1 5 π 4 5 1 e 18..6 It is easy to check that for max {[ ] } 490 α +, 5505, α 1 5 π 4 5 1 e 18 > ad that for max{[α + 4 ] +, 5505}, Let α 1 10470 1 4 > 0, 4 log[µ 1] α log 1 6 > 1 1 1 α /8 1 > 1 11/4 1. {[ ] 490 α = max α It ca be see that for > α, } +, [α + 4 ] +, 5505. 4 log[µ 1] α log 1 + α 1 1 1 5 π 4 5 1 1 e 18 >..6 1 Combig.57 ad.6, we deduce that the right had side of.6 is positive for > α. So we are led to the log-covexity of the sequece { p/ α } α. 1 is
A iequality o the ratio 1 p 1 p I this sectio, we employ Lemma. ad Lemma. to fid the limit of 5 log p. The we give a upper boud for log 1 p 1. This leads to the iequality 1.1. Theorem.1 Let = π/ 4. We have lim + 5 log p =..1 Proof. Usig.8, that is, the N = case of the Hardy-Ramauja-Rademacher formula for p, we fid that log p = 1 log T + 1 log1 + ỹ, where T ad y are give by.9 ad.1. By the defiitio.14 of Ẽ, we get log 1 p 1 = 1 1 log T 1 + Ẽ 1.. Applyig Lemma., we get that From Lemma., we get Usig.,. ad.4, we deduce that as required. lim 1 5 1 + 1 log T 1 =.. lim 1 5 Ẽ 1 = 0..4 + lim 5 log p =, + To prove Theorem 1.4, we eed the followig upper boud for log 1 p 1. Theorem. For, log 1 p 1 < π 4 + π..5 14
Proof. By the upper boud of 1 1 log T 1 give i Lemma., the upper boud of Ẽ 1 give i Lemma., ad the relatio., we obtai the followig upper boud of log 1 p 1 for 40: log 1 p 1 < To prove.5, we claim that for 095, 14 5 + 5 log[µ + 1] 4 + 5 π 4 5 / 1 + 1 1 e 18. 14 5 + 5 4 log[µ + 1] + 5 π 4 5 / 1 + 1 1 e 18 < First, we show that for 60, π 4 + π..6 14 5 π / 4 + π < 1 1..7 For 0 < x 1, it ca be checked that 48 I the otatio = π/ 4, we have Settig x = 5 1, we have x 4.9, we fid that for 60, so that for 60, 1 1 x / < 1 + x + 8 x..8 14 5 = / 1 / 1 5..9 / 4 48 1 / 1 5 4 / < 1[4 5] π / 4 + π < 1 π / 4 + π + for 60. Applyig.8 to the right had side of [ 1 + 75 1 / 48 + 5 8 4 [ 1 / 75 48 + 8 5 4 ],.10 ]..11 To prove.7, we proceed to show that the right had side of.11 is bouded by 1 1. Notig that for, 1 π / 4 + π = + 1 + + 1 /, 15
ad usig the fact + > 1, together with / > 1 /, we deduce that 1 / π 4 + π < Applyig.1 to.11, we obtai that for 60, 1[4 5] / π 4 + π < [ 1 + 7/ 1 + 5 1 / 1 7/ + 1 5..1 75 48 + 8 5 4 ]..1 Sice 75 < ad 5 1 < for, it follows from.1 that for 60, 48 1 8 4 1 / 1[4 5] π / 4 + π Usig the fact that <, we see that < 1 7/ + 1 5 + 1 7/ + 1 4. 1 7/ + 1 5 + 1 4 < 6 1 7/ + 4 1 5 + 1 4..14 For 60, it is easily checked that the right had side of.14 is bouded by 1 1. This cofirms.7. To prove the claim.6, it is eough to show that for 095, 1 4 log[µ + 1] 5 < 1 + 1 1 5 π 4 5 1 e 18..15 From.61 it ca be see that for 095, 5 π 4 5 1 e 18 < 5 1..16 Sice 4 log[µ + 1] > 18 for 095, it follows from.16 that for 095, 4 log[µ + 1] 5 + 1 1 5 π 4 5 1 e 18 > 18 + 1 10 1 > 1 1. So we obtai.15. Combiig.15 ad??, we arrive at.6. For 094, the iequality.5 ca be easily checked. This completes the proof. 16
We are ow i a positio to complete the proof of Theorem 1.4. Proof of Theorem 1.4. It is kow that for x > 0, so that for 1, x 1 + x < log1 + x, π 1 4 + π < log π +. 4 I light of the above relatio, Theorem. implies that for, log 1 π p 1 < log 1 +, 4 that is, +1 p + 1 1 p 1 < π 1 + p, 4 as required. We remark that = π/ 4 is the smallest possible umber for the iequality i Theorem 1.4. Suppose that 0 < γ <. By Theorem.1, there exists a iteger N such that for > N, log 1 p 1 > γ. It follows that log 1 p 1 > which implies that for > N, p +1 p + 1 γ > log 1 + γ, 1 + γ < 1 p 1. p Ackowledgmets. We wish to thak the referee for helpful commets. Refereces [1] G.E. Adrews, Combiatorial proof of a partitio fuctio limit, Amer. Math. Mothly., 78 1971, 76 78. [] G.E. Adrews, The Theory of Partitios, Cambridge Uiversity Press, Cambridge, 1998. [] W.Y.C. Che, J.J.F. Guo ad L.X.W. Wag, Ifiitely log-mootoic combiatorial sequeces. Adv. Appl. Math., 5 014, 99 10. 17
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