CUBO A Mathemati al Journal Vol.15, N o 03, (105�122). O tober 2013 Euler's onstant, new lasses of sequen es and estimates Alina Sînt m rian Department of Mathemati s, Te hni al University of Cluj-Napo a, Str. Memorandumului nr. 28, 400114 Cluj-Napo a, Romania. alina.sintamarian�math.ut luj.ro ABSTRACT We give two lasses of sequen es with the argument of the logarithmi term modi- �ed and also with some additional terms besides those in the de�nition sequen e, and that onverge qui kly to γ(a) = lim n→∞ (∑n k=1 1 a+k−1 − ln a+n−1 a ) , where a ∈ (0,+∞). We present the pattern in forming these sequen es, expressing the oe� ients that appear with the Bernoulli numbers. Also, we obtain estimates ontaining best on- stants for ∑n k=1 1 k − 1 24(n+1/2)2 − ln ( n + 1 2 − 7 960(n+1/2)3 ) −γ and γ− (∑n k=1 1 k − 1 24(n+1/2)2 + 7 960(n+1/2)4 − ln ( n + 1 2 + 31 8064(n+1/2)5 )) , where γ = γ(1) is the Euler's onstant. RESUMEN Mostramos dos lases de se uen ias on el argumento del término logarítmi o modi�- ado y también on algunos términos adi ionales además de los de�nidos en la se uen- ia y que onvergen rápidamente a γ(a) = lim n→∞ (∑n k=1 1 a+k−1 − ln a+n−1 a ) , donde a ∈ (0,+∞). Presentamos el patrón que forma las se uen ias expresando los oe� ientes que apare en en los números de Bernoulli. Además, obtenemos estima iones que on- tienen las mejores onstantes para ∑n k=1 1 k − 1 24(n+1/2)2 −ln ( n + 1 2 − 7 960(n+1/2)3 ) −γ y γ − (∑n k=1 1 k − 1 24(n+1/2)2 + 7 960(n+1/2)4 − ln ( n + 1 2 + 31 8064(n+1/2)5 )) , donde γ = γ(1) es la onstante de Euler. Keywords and Phrases: sequen e, onvergen e, approximation, Euler's onstant, Bernoulli number, estimate. 2010 AMS Mathemati s Subje t Classi� ation: 11Y60, 11B68, 40A05, 41A44, 33B15. 106 Alina Sînt m rian CUBO 15, 3 (2013) 1 Introdu tion Let Hn = ∑n k=1 1/k be the nth harmoni number and let Dn = Hn − lnn. Euler's onstant γ = limn→∞ Dn is one of the most important onstants in mathemati s and is also the topi of many papers in the literature. This omes as a on�rmation 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 2n + 2γ−1 1−γ ≤ Dn − γ < 1 2n + 1 3 , n ∈ N, the numbers 2γ−1 1−γ and 1 3 being the best onstants with this property, i.e. 2γ−1 1−γ annot be repla ed by a smaller one and 1 3 annot be repla ed 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 sequen e (Dn)n∈N onverges to γ very slowly. In order to in rease the speed of onvergen e to γ, D. W. DeTemple [7℄ modi�ed the argument of the logarithmi term from Dn, onsidering the sequen e (Rn)n∈N de�ned by Rn = Hn − ln(n+1/2), and he proved that 1 24(n+1)2 < Rn −γ < 1 24n2 , n ∈ N. Sequen es with higher rate of onvergen e to γ an be also obtained by subtra ting a rational term from Dn: it is shown (see [21℄) that limn→∞ n 2(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 subtra ting and adding terms in the de�nition sequen e, to obtain qui ker onvergen es 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 12n2+a2 < γ − ( Dn − 1 2n ) ≤ 1 12n2+b2 , n ∈ N ([9℄); 7 960(n+a3) 4 ≤ γ − ( Hn − ln ( n + 1 2 ) − 1 24(n+1/2)2 ) < 7 960(n+b3) 4 , n ∈ N ([4℄); 17 3840(n+a4) 5 ≤ Hn − ln ( n + 1 2 + 1 24n − 1 48n2 + 23 5760n3 ) − γ < 17 3840(n+b4) 5 , n ∈ N ([6℄), with a1 = 1/ √ 24(1 − γ − ln(3/2)) − 1 and b1 = 1/2; a2 = 6/5 and b2 = 2(7 − 12γ)/(2γ − 1); a3 = 1/ 4 √ 960/7(ln(3/2) + γ − 53/54) − 1 and b3 = 1/2; 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 anti ipated earlier, in the present paper we shall investigate a generalization of Euler's onstant, namely the limit γ(a) of the sequen e (yn(a))n∈N de�ned by (see [11, p. 453℄, [16℄, [17℄, CUBO 15, 3 (2013) Euler's onstant, new lasses of sequen es and estimates 107 [18℄) yn(a) = n∑ k=1 1 a + k − 1 − ln a + n − 1 a , 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 Se tion 2 we give two lasses of sequen es with the argument of the logarithmi term modi�ed and also with some additional terms besides those in the de�nition sequen e (yn(a))n∈N, and that onverge qui kly to γ(a). We present the pattern in forming these sequen es, expressing the oe� ients that appear with the Bernoulli numbers. The two lasses of sequen es have as base the sequen es (αn,2(a))n∈N and (αn,3(a))n∈N de�ned by αn,2(a) = n∑ k=1 1 a + k − 1 − 1 24 ( a + n − 1 2 )2 − ln ( a + n − 1 2 a − 7 960a ( a + n − 1 2 )3 ) , αn,3(a) = n∑ k=1 1 a + k − 1 − 1 24 ( a + n − 1 2 )2 + 7 960 ( a + n − 1 2 )4 − ln ( a + n − 1 2 a + 31 8064a ( a + n − 1 2 )5 ) . In Se tion 3 we prove estimates ontaining best onstants for αn,2(1) −γ and γ −αn,3(1), n ∈ N. The following lemma, whi h we shall need in our proofs, was given by C. Morti i [13, Lemma℄ and is a onsequen e 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 sequen e of real numbers and x ∗ = lim n→∞ xn. We suppose that there exists α ∈ R, α > 1, su h that lim n→∞ nα(xn − xn+1) = l ∈ R. Then there exists the limit lim n→∞ nα−1(xn − x ∗ ) = l α − 1 . Also, re all that the digamma fun tion ψ is the logarithmi derivative of the gamma fun tion, 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. (1) From the re urren e formula ( [1, p. 258℄) ψ(x + 1) = ψ(x) + 1 x , x ∈ (0,+∞), 108 Alina Sînt m rian CUBO 15, 3 (2013) and the asymptoti formula ( [1, p. 259℄) ψ(x) ∼ lnx − 1 2x − 1 12x2 + 1 120x4 − 1 252x6 + 1 240x8 − 1 132x10 + · · · (x → ∞), one obtains ψ(x + 1) ∼ lnx + 1 2x − 1 12x2 + 1 120x4 − 1 252x6 + 1 240x8 − 1 132x10 + · · · (x → ∞). (2) 2 Sequen es that onverge to γ(a) Theorem 2.1. Let a ∈ (0,+∞) and let γ(a) be the limit of the sequen e (yn(a))n∈N from Intro- du tion. We onsider the sequen es (αn,2(a))n∈N and (βn,2(a))n∈N de�ned by αn,2(a) = n∑ k=1 1 a + k − 1 − 1 24 ( a + n − 1 2 )2 − ln ( a + n − 1 2 a − 7 960a ( a + n − 1 2 )3 ) , βn,2(a) = αn,2(a) − 31 8064 ( a + n − 1 2 )6 . Then: (i) lim n→∞ n6(αn,2(a) − γ(a)) = 31 8064 ; (ii) lim n→∞ n8(γ(a) − βn,2(a)) = 7571 1843200 . Proof. (i) Set εn := 1 a+n , n ∈ N. Sin e ±1 2 εn ∈ (−1,1), − 1 2 εn − 7 960 · ε 4 n (1− 12 εn) 3 ∈ (−1,1] and 1 2 εn − 7 960 · ε 4 n (1+ 12 εn) 3 ∈ (−1,1], for every n ∈ N, using the series expansion ( [11, pp. 171�179℄) we obtain αn,2(a) − αn+1,2(a) = −εn − 1 24 · ε2n ( 1 − 1 2 εn )2 + 1 24 · ε2n ( 1 + 1 2 εn )2 − ln ( 1 − 1 2 εn − 7 960 · ε4n ( 1 − 1 2 εn )3 ) + ln ( 1 + 1 2 εn − 7 960 · ε4n ( 1 + 1 2 εn )3 ) = 31 1344 ε7n + 4829 230400 ε9n + 2913 225280 ε11n + 20456239 2875392000 ε13n + O(ε 15 n ). It follows that lim n→∞ n7(αn,2(a) − αn+1,2(a)) = 31 1344 . Now, a ording to Lemma 1.1, we get lim n→∞ n6(αn,2(a) − γ(a)) = 31 8064 . CUBO 15, 3 (2013) Euler's onstant, new lasses of sequen es and estimates 109 (ii) We are able to write that βn+1,2(a) − βn,2(a) = αn+1,2(a) − αn,2(a) − 31 8064 · ε6n (1 + 1 2 εn) 6 + 31 8064 · ε6n (1 − 1 2 εn) 6 = 7571 230400 ε9n + 10727 225280 ε11n + O(ε 13 n ). Therefore lim n→∞ n9(βn+1,2(a) − βn,2(a)) = 7571 230400 , and based on Lemma 1.1 we obtain lim n→∞ n8(γ(a) − βn,2(a)) = 7571 1843200 . Also, onsidering the sequen e in ea h of the following parts and using similar arguments as in Theorem 2.1, we get the indi ated limit: δn,2(a) = βn,2(a) + 7571 1843200 ( a + n − 1 2 )8 , n ∈ N, lim n→∞ n10(δn,2(a) − γ(a)) = 511 67584 ; ηn,2(a) = δn,2(a) − 511 67584 ( a + n − 1 2 )10 , n ∈ N, lim n→∞ n12(γ(a) − ηn,2(a)) = 5092085987 241532928000 ; θn,2(a) = ηn,2(a) + 5092085987 241532928000 ( a + n − 1 2 )12 , n ∈ N, lim n→∞ n14(θn,2(a) − γ(a)) = 8191 98304 ; λn,2(a) = θn,2(a) − 8191 98304 ( a + n − 1 2 )14 , n ∈ N, lim n→∞ n16(γ(a) − λn,2(a)) = 25599939583183 57755566080000 ; µn,2(a) = λn,2(a) + 25599939583183 57755566080000 ( a + n − 1 2 )16 , n ∈ N, lim n→∞ n18(µn,2(a) − γ(a)) = 5749691557 1882718208 . 110 Alina Sînt m rian CUBO 15, 3 (2013) We remark the pattern in forming the sequen es from Theorem 2.1 and those mentioned above. For example, the general term of the sequen e (µn,2(a))n∈N an be written in the form µn,2(a) = n∑ k=1 1 a + k − 1 − 1 2 · B2 2 · 1 ( a + n − 1 2 )2 − ln ( a + n − 1 2 a + 23 − 1 23 · B4 4 · 1 a ( a + n − 1 2 )3 ) − 8∑ k=3 ck,2 ( a + n − 1 2 )2k , with ck,2 =    22k−1 − 1 22k−1 · B2k 2k , if k = 2p + 1,p ∈ N, 22k−1 − 1 22k−1 · B2k 2k + 2 k ( − 23 − 1 23 · B4 4 ) k 2 , if k = 2p + 2,p ∈ N, 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 a urate to nine de imal pla es in approximating γ. Theorem 2.2. Let a ∈ (0,+∞) and let γ(a) be the limit of the sequen e (yn(a))n∈N from Intro- du tion. We onsider the sequen es (αn,3(a))n∈N, (βn,3(a))n∈N and (δn,3(a))n∈N de�ned by αn,3(a) = n∑ k=1 1 a + k − 1 − 1 24 ( a + n − 1 2 )2 + 7 960 ( a + n − 1 2 )4 − ln ( a + n − 1 2 a + 31 8064a ( a + n − 1 2 )5 ) , βn,3(a) = αn,3(a) + 127 30720 ( a + n − 1 2 )8 , δn,3(a) = βn,3(a) − 511 67584 ( a + n − 1 2 )10 . Then: (i) lim n→∞ n8(γ(a) − αn,3(a)) = 127 30720 ; CUBO 15, 3 (2013) Euler's onstant, new lasses of sequen es and estimates 111 (ii) lim n→∞ n10(βn,3(a) − γ(a)) = 511 67584 ; (iii) lim n→∞ n12(γ(a) − δn,3(a)) = 178161637 8453652480 . Proof. (i) Set εn := 1 a+n , n ∈ N. Sin e ±1 2 εn ∈ (−1,1), 1 2 εn + 31 8064 · ε 6 n (1+ 12 εn) 5 ∈ (−1,1] and −1 2 εn + 31 8064 · ε 6 n (1− 12 εn) 5 ∈ (−1,1], for every n ∈ N, using the series expansion ( [11, pp. 171�179℄) we obtain αn+1,3(a) − αn,3(a) = εn − 1 24 · ε2n ( 1 + 1 2 εn )2 + 1 24 · ε2n ( 1 − 1 2 εn )2 + 7 960 · ε4n ( 1 + 1 2 εn )4 − 7 960 · ε4n ( 1 − 1 2 εn )4 − ln ( 1 + 1 2 εn + 31 8064 · ε6n ( 1 + 1 2 εn )5 ) + ln ( 1 − 1 2 εn + 31 8064 · ε6n ( 1 − 1 2 εn )5 ) = 127 3840 ε9n + 409 8448 ε11n + 5873471 140894208 ε13n + 2502391 92897280 ε15n + 2826605 210567168 ε17n + 33340423721 8302787297280 ε19n + O(ε 21 n ). Consequently, lim n→∞ n9(αn+1,3(a) − αn,3(a)) = 127 3840 , and from this, based on Lemma 1.1, we get lim n→∞ n8(γ(a) − αn,3(a)) = 127 30720 . (ii) We have βn,3(a) − βn+1,3(a) = αn,3(a) − αn+1,3(a) + 127 30720 · ε8n (1 − 1 2 εn) 8 − 127 30720 · ε8n (1 + 1 2 εn) 8 = 2555 33792 ε11n + 114794663 704471040 ε13n + 18092183 92897280 ε15n + 36074257 210567168 ε17n + O(ε 19 n ). Thus lim n→∞ n11(βn,3(a) − βn+1,3(a)) = 2555 33792 , and applying Lemma 1.1, it follows that lim n→∞ n10(βn,3(a) − γ(a)) = 511 67584 . 112 Alina Sînt m rian CUBO 15, 3 (2013) (iii) We an write that δn+1,3(a) − δn,3(a) = βn+1,3(a) − βn,3(a) − 511 67584 · ε10n (1 + 1 2 εn) 10 + 511 67584 · ε10n (1 − 1 2 εn) 10 = 178161637 704471040 ε13n + 69794707 92897280 ε15n + O(ε 17 n ). Hen e lim n→∞ n13(δn+1,3(a) − δn,3(a)) = 178161637 704471040 . This, along with Lemma 1.1, gives lim n→∞ n12(γ(a) − δn,3(a)) = 178161637 8453652480 . Also, onsidering the sequen e in ea h of the following parts and using similar arguments as in Theorem 2.2, we get the indi ated limit: ηn,3(a) = δn,3(a) + 178161637 8453652480 ( a + n − 1 2 )12 , n ∈ N, lim n→∞ n14(ηn,3(a) − γ(a)) = 8191 98304 ; θn,3(a) = ηn,3(a) − 8191 98304 ( a + n − 1 2 )14 , n ∈ N, lim n→∞ n16(γ(a) − θn,3(a)) = 118518239 267386880 ; λn,3(a) = θn,3(a) + 118518239 267386880 ( a + n − 1 2 )16 , n ∈ N, lim n→∞ n18(λn,3(a) − γ(a)) = 91282102592903 29890034270208 ; µn,3(a) = λn,3(a) − 91282102592903 29890034270208 ( a + n − 1 2 )18 , n ∈ N, lim n→∞ n20(γ(a) − µn,3(a)) = 91546277357 3460300800 . We remark the pattern in forming the sequen es from Theorem 2.2 and those mentioned above. For example, the general term of the sequen e (µn,3(a))n∈N an be written in the form µn,3(a) = n∑ k=1 1 a + k − 1 − 1 2 · B2 2 · 1 ( a + n − 1 2 )2 − 23 − 1 23 · B4 4 · 1 ( a + n − 1 2 )4 − ln ( a + n − 1 2 a + 25 − 1 25 · B6 6 · 1 a ( a + n − 1 2 )5 ) − 9∑ k=4 ck,3 ( a + n − 1 2 )2k , CUBO 15, 3 (2013) Euler's onstant, new lasses of sequen es and estimates 113 with ck,3 =    22k−1 − 1 22k−1 · B2k 2k , if k = 3p + 1,p ∈ N, 22k−1 − 1 22k−1 · B2k 2k , if k = 3p + 2,p ∈ N, 22k−1 − 1 22k−1 · B2k 2k + 3 k ( − 25 − 1 25 · B6 6 ) k 3 , if k = 3p + 3,p ∈ N, 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 a urate to nine de imal pla es in approximating γ. Con luding, the following remark an be made. Let a ∈ (0,+∞) and q ∈ N \ {1}. Let n0 = min { n ∈ N ∣ ∣ ∣ ∣ a + n − 1 2 + 2 2q−1 −1 22q−1 · B2q 2q · 1 (a+n− 12) 2q−1 > 0 } . We onsider the sequen e (αn,q(a))n≥n0 de�ned by αn,q(a) = n∑ k=1 1 a + k − 1 − q−1∑ k=1 22k−1 − 1 22k−1 · B2k 2k · 1 ( a + n − 1 2 )2k − ln ( a + n − 1 2 a + 22q−1 − 1 22q−1 · B2q 2q · 1 a ( a + n − 1 2 )2q−1 ) , for every n ∈ N, n ≥ n0. Clearly, lim n→∞ αn,q(a) = γ(a). Based on the sequen e (αn,q(a))n≥n0, a lass of sequen es onvergent to γ(a) an be onsidered, namely {(αn,q,r(a))n≥n0|r ∈ N,r ≥ q+1}, where αn,q,r(a) = αn,q(a) − r∑ k=q+1 ck,q ( a + n − 1 2 )2k , for every n ∈ N, n ≥ n0, with ck,q =    22k−1 − 1 22k−1 · B2k 2k , if k ∈ {qp + 1,qp + 2, . . . ,qp + q − 1},p ∈ N, 22k−1 − 1 22k−1 · B2k 2k + q k ( − 22q−1 − 1 22q−1 · B2q 2q ) k q , if k = qp + q,p ∈ N. 114 Alina Sînt m rian CUBO 15, 3 (2013) In this se tion we have obtained that the sequen e (αn,q(a))n∈N onverges to γ(a) as n −(2q+2) and that the sequen e (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 sequen e de�ned by αn = αn,2(1). In part (i) of Theorem 2.1 we have proved that lim n→∞ n6(αn − γ) = 31 8064 . (3) Proposition 3.1. We have γ < αn+1 < αn, for every n ∈ N. Proof. We have αn+1 − αn = 1 n + 1 − 1 24 ( n + 3 2 )2 + 1 24 ( n + 1 2 )2 − ln 960 ( n + 3 2 )4 − 7 ( n + 3 2 )3 + ln 960 ( n + 1 2 )4 − 7 ( n + 1 2 )3 . Considering the fun tion h : [1,+∞) → R, de�ned by h(x) = 1 x + 1 − 1 24 ( x + 3 2 )2 + 1 24 ( x + 1 2 )2 − ln 960 ( x + 3 2 )4 − 7 ( x + 3 2 )3 + ln 960 ( x + 1 2 )4 − 7 ( x + 1 2 )3 , and di�erentiating it, we obtain that h′(x) = − 1 (x + 1)2 + 1 12 ( x + 3 2 )3 − 1 12 ( x + 1 2 )3 − 3840 ( x + 3 2 )3 960 ( x + 3 2 )4 − 7 + 3 x + 3 2 + 3840 ( x + 1 2 )3 960 ( x + 1 2 )4 − 7 − 3 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 fun tion h is stri tly in reasing 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 sequen e (αn)n∈N is stri tly de reasing. Be ause limn→∞ αn = γ, we on lude that γ < αn+1 < αn, for every n ∈ N. Now we give our �rst main result of this se tion. CUBO 15, 3 (2013) Euler's onstant, new lasses of sequen es and estimates 115 Theorem 3.2. Let c = 6 √ 31 8064( 5354 −ln 4853 3240 −γ) . We have 31 8064(n + c − 1)6 ≤ αn − γ < 31 8064 ( n + 1 2 )6 , 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 fun tion from the proof of Proposition 3.1. Let (un)n∈N be the sequen e de�ned by un = αn − 31 8064(n + c − 1)6 . We have un+1 − un = αn+1 − αn − 31 8064(n + c)6 + 31 8064(n + c − 1)6 . We onsider the fun tion f : [1,+∞) → R de�ned by f(x) = h(x) − 31 8064(x + c)6 + 31 8064(x + c − 1)6 . Di�erentiating, we get that f′(x) = h′(x) + 31 1344(x + c)7 − 31 1344(x + c − 1)7 = [(x − 2) 20∑ k=0 akx k + a]/[1344(x + 1)2(2x + 1)3(2x + 3)3 ×(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,+∞). Hen e, the fun tion f is stri tly in reasing 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 sequen e (un)n≥2 is stri tly de reasing. Having in view that limn→∞ un = γ, we are able to write that γ < un, for every n ≥ 2. Consequently, 31 8064(n + c − 1)6 ≤ αn − γ, 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 sequen e de�ned by vn = αn − 31 8064 ( n + 1 2 )6 . 116 Alina Sînt m rian CUBO 15, 3 (2013) Then vn+1 − vn = αn+1 − αn − 31 8064 ( n + 3 2 )6 + 31 8064 ( n + 1 2 )6 . Di�erentiating the fun tion g : [1,+∞) → R, de�ned by g(x) = h(x) − 31 8064 ( x + 3 2 )6 + 31 8064 ( x + 1 2 )6 , we obtain that g′(x) = h′(x) + 31 1344 ( x + 3 2 )7 − 31 1344 ( x + 1 2 )7 = −(93776707584x14 + 1312873906176x13 + 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 fun tion g is stri tly de reasing 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, whi h means that the sequen e (vn)n∈N is stri tly in reasing. Sin e limn→∞ vn = γ, it follows that vn < γ, for every n ∈ N. We an therefore write that αn − γ < 31 8064 ( n + 1 2 )6 , (4) for every n ∈ N. It remains to prove that the onstant 1 2 is the best possible with the property that the above inequality (4) holds for every n ∈ N, and this an be a hieved as follows. We have just proved that 6 √ 31 8064(αn − γ) − n > 1 2 , n ∈ N. (5) CUBO 15, 3 (2013) Euler's onstant, new lasses of sequen es and estimates 117 Using (1) and (2), we get that αn − γ = Hn − 1 24 ( n + 1 2 )2 − ln ( n + 1 2 − 7 960 ( n + 1 2 )3 ) − γ = ψ(n + 1) − 1 24 ( n + 1 2 )2 − ln ( n + 1 2 − 7 960 ( n + 1 2 )3 ) = 1 2n − 1 12n2 + 1 120n4 − 1 252n6 + 1 240n8 + O ( 1 n10 ) − 1 24n2 ( 1 + 1 2n )2 − ln ( 1 + 1 2n − 7 960n4 ( 1 + 1 2n )3 ) = 31 8064n6 − 31 2688n7 + O ( 1 n8 ) . (6) Let An = 6 √ 31 8064n6(αn−γ) , n ∈ N. Clearly, lim n→∞ An = 1, having in view (3). Then, based on (6), we have 6 √ 31 8064(αn − γ) − n = n(An − 1) = n ∑5 k=0 Akn ( 1 8064 31 n6(αn − γ) − 1 ) = n ∑5 k=0 Akn ( 1 1 − 3 n + O ( 1 n2 ) − 1 ) = 1 ∑5 k=0 Akn · 3 + O ( 1 n ) 1 − 3 n + O ( 1 n2 ) → 1 6 · 3 = 1 2 (n → ∞). (7) Indeed, from (5) and (7) we obtain that 1 2 is the best onstant with the property that inequality (4) holds for every n ∈ N, and now the proof is omplete. Let (�αn)n∈N be the sequen e de�ned by �αn = αn,3(1). In part (i) of Theorem 2.2 we have proved that lim n→∞ n8(γ − �αn) = 127 30720 . (8) Proposition 3.3. We have �αn < �αn+1 < γ, for every n ∈ N. Proof. We have �αn+1 − �αn = 1 n + 1 − 1 24 ( n + 3 2 )2 + 1 24 ( n + 1 2 )2 + 7 960 ( n + 3 2 )4 − 7 960 ( n + 1 2 )4 − ln 8064 ( n + 3 2 )6 + 31 ( n + 3 2 )5 + ln 8064 ( n + 1 2 )6 + 31 ( n + 1 2 )5 . 118 Alina Sînt m rian CUBO 15, 3 (2013) Considering the fun tion �h : [1,+∞) → R, de�ned by �h(x) = 1 x + 1 − 1 24 ( x + 3 2 )2 + 1 24 ( x + 1 2 )2 + 7 960 ( x + 3 2 )4 − 7 960 ( x + 1 2 )4 − ln 8064 ( x + 3 2 )6 + 31 ( x + 3 2 )5 + ln 8064 ( x + 1 2 )6 + 31 ( x + 1 2 )5 , and di�erentiating it, we obtain that �h′(x) = − 1 (x + 1)2 + 1 12 ( x + 3 2 )3 − 1 12 ( x + 1 2 )3 − 7 240 ( x + 3 2 )5 + 7 240 ( x + 1 2 )5 − 48384 ( x + 3 2 )5 8064 ( x + 3 2 )6 + 31 + 5 x + 3 2 + 48384 ( x + 1 2 )5 8064 ( x + 1 2 )6 + 31 − 5 x + 1 2 = −(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, for every x ∈ [1,+∞). It follows that the fun tion �h is stri tly de reasing 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 sequen e (�αn)n∈N is stri tly in reasing. Be ause limn→∞ �αn = γ, we on lude that �αn < �αn+1 < γ, for every n ∈ N. Now we give our se ond main result of this se tion. Theorem 3.4. Let �c = 8 √ 127 30720(γ− 47774860 −ln 91885 61236) . We have 127 30720(n + �c − 1)8 ≤ γ − �αn < 127 30720 ( n + 1 2 )8 , 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 fun tion from the proof of Proposition 3.3. Let (�un)n∈N be the sequen e de�ned by �un = �αn + 127 30720(n + �c − 1)8 . CUBO 15, 3 (2013) Euler's onstant, new lasses of sequen es and estimates 119 We have �un+1 − �un = �αn+1 − �αn + 127 30720(n + �c)8 − 127 30720(n + �c − 1)8 . We onsider the fun tion �f : [1,+∞) → R de�ned by �f(x) = �h(x) + 127 30720(x + �c)8 − 127 30720(x + �c − 1)8 . Di�erentiating, we get that �f′(x) = �h′(x) − 127 3840(x + �c)9 + 127 3840(x + �c − 1)9 = −[(x − 2) 30∑ k=0 �akx k + �a]/[3840(x + 1)2(2x + 1)5(2x + 3)5 ×(8064x6 + 24192x5 + 30240x4 + 20160x3 + 7560x2 + 1512x + 157) ×(8064x6 + 72576x5 + 272160x4 + 544320x3 + 612360x2 +367416x + 91885)(x + �c)9(x + �c − 1)9]. One an verify that �ai > 0, i ∈ {0,1, . . . ,30} and �a > 0. It follows that �f ′(x) < 0, for every x ∈ [2,+∞). Hen e, the fun tion �f is stri tly de reasing 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 sequen e (�un)n≥2 is stri tly in reasing. Having in view that limn→∞ �un = γ, we are able to write that �un < γ, for every n ≥ 2. Consequently, 127 30720(n + �c − 1)8 ≤ γ − �αn, 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 sequen e de�ned by �vn = �αn + 127 30720 ( n + 1 2 )8 . Then �vn+1 − �vn = �αn+1 − �αn + 127 30720 ( n + 3 2 )8 − 127 30720 ( n + 1 2 )8 . Di�erentiating the fun tion �g : [1,+∞) → R, de�ned by �g(x) = �h(x) + 127 30720 ( x + 3 2 )8 − 127 30720 ( x + 1 2 )8 , 120 Alina Sînt m rian CUBO 15, 3 (2013) we obtain that �g′(x) = �h′(x) − 127 3840 ( x + 3 2 )9 + 127 3840 ( x + 1 2 )9 = (212667885158400x20 + 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)]. Thus �g′(x) > 0, for every x ∈ [1,+∞). Hereby, the fun tion �g is stri tly in reasing 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, whi h means that the sequen e (�vn)n∈N is stri tly de reasing. Sin e limn→∞ �vn = γ, it follows that γ < �vn, for every n ∈ N. We an therefore write that γ − �αn < 127 30720 ( n + 1 2 )8 , (9) for every n ∈ N. It remains to prove that the onstant 1 2 is the best possible with the property that the above inequality (9) holds for every n ∈ N, and this an be a hieved as follows. We have just proved that 8 √ 127 30720(γ − �αn) − n > 1 2 , n ∈ N. (10) Using (1) and (2), we get that γ − �αn = γ − Hn + 1 24 ( n + 1 2 )2 − 7 960 ( n + 1 2 )4 + ln ( n + 1 2 + 31 8064 ( n + 1 2 )5 ) = −ψ(n + 1) + 1 24 ( n + 1 2 )2 − 7 960 ( n + 1 2 )4 + ln ( n + 1 2 + 31 8064 ( n + 1 2 )5 ) = − 1 2n + 1 12n2 − 1 120n4 + 1 252n6 − 1 240n8 + 1 132n10 + O ( 1 n12 ) + 1 24n2 ( 1 + 1 2n )2 − 7 960n4 ( 1 + 1 2n )4 + ln ( 1 + 1 2n + 31 8064n6 ( 1 + 1 2n )5 ) = 127 30720n8 − 127 7680n9 + O ( 1 n10 ) . (11) CUBO 15, 3 (2013) Euler's onstant, new lasses of sequen es and estimates 121 Let �An = 8 √ 127 30720n8(γ−�αn) , n ∈ N. Clearly, lim n→∞ �An = 1, having in view (8). Then, based on (11), we have 8 √ 127 30720(γ − �αn) − n = n(�An − 1) = n ∑7 k=0 �Akn ( 1 30720 127 n8(γ − �αn) − 1 ) = n ∑7 k=0 �Akn ( 1 1 − 4 n + O ( 1 n2 ) − 1 ) = 1 ∑7 k=0 �Akn · 4 + O ( 1 n ) 1 − 4 n + O ( 1 n2 ) → 1 8 · 4 = 1 2 (n → ∞). (12) Combining (10) and (12) we obtain that 1 2 is the best onstant with the property that inequality (9) holds for every n ∈ N, and now the proof is omplete. Re eived: September 2012. A epted: September 2013. Referen es [1℄ M. Abramowitz, I. A. Stegun, Handbook of Mathemati al Fun tions with Formulas, Graphs, and Mathemati al Tables, National Bureau of Standards Applied Mathemati s Series 55, Washington, 1964. [2℄ H. Alzer, Inequalities for the gamma and polygamma fun tions, Abh. Math. Semin. Univ. Hamb. 68, 1998, 363�372. [3℄ C.-P. 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