《欧拉常数》（英文版）阅读材料——欧拉常数

ON EULER'S CONSTANT -CALCULATING SUMS BY INTEGRALS LI YINGYING Euler's constant is defined as y= lim Dn, (1) n→∞ where Dn=∑-log(n+1)(n∈n) (2) Write =y-Dn. (3) R.M. Young [1] gave an estimate of Tn: 1 1 2(n+1) -<< 2n (4 D.W. DeTemple [2] considered n k=1 instead of Dn and showed 71 1 960(n+1)4-Dn+ 24n+ An earlier discussion on Dn can be found in Rippon 3]. Furthermore, DeTemple and Wang [4] established In this article we give an exact expansion of using totally elementary method from which asymptotic estimates for Tn are derived. Our method is to calculate sums by integrals. Rewrite Dn a n as Dn= tdt. k(k+t) (*) Key words and phrases. Euler constant, sums, integrals. Mathematics Subject Classification 2000, 40A05 The project was supported by a grant from the Education Ministry of China Typeset by AMS-TEX

ON EULER’S CONSTANT —CALCULATING SUMS BY INTEGRALS Li Yingying Euler’s constant is defined as γ = limn→∞ Dn, (1) where Dn = Xn k=1 1 k − log(n + 1) (n ∈ N). (2) Write rn = γ − Dn. (3) R.M. Young [1] gave an estimate of rn: 1 2(n + 1) < rn < 1 2n . (4) D.W. DeTemple [2] considered D˜ n := Xn k=1 1 k − log ³ n + 1 2 ´ instead of Dn and showed 7 960 · 1 (n + 1)4 < γ − D˜ n + 1 24³ n + 1 2 ´2 < 7 960n4 , An earlier discussion on Dn can be found in Rippon [3]. Furthermore, DeTemple and Wang [4] established an estimate for rn with arbitrary n where Bernoulli’s numbers are involved. In this article we give an exact expansion of rn using totally elementary method from which asymptotic estimates for rn are derived. Our method is to calculate sums by integrals. Rewrite Dn as Dn = Xn k=1 Ã 1 k − Z k+1 k 1 x dx! = Xn k=1 Z 1 0 t k(k + t) dt. (∗) Key words and phrases. Euler constant, sums, integrals. Mathematics Subject Classification 2000, 40A05 The project was supported by a grant from the Education Ministry of China Typeset by AMS-TEX 1

LI YINGYING In= ∑、/(a dt+ k=n+1 t)k(k+1) k=n+1A(k+1) t(1-t) k=n+ k(k+1)(k+t Then rn=ri(n)+ n+1 k(k+1)(k+t) 1 1 t(1-t) k(k+1)(k+t)k(k+1)(k+2) t(1-t) k(k+1)(k+2) t(1-t)(2-t) k(k+1)(k+2)(k+t) 2 ta-ddt(a k=n+ k(k+1)(k+1)(k+2) dt+ k(k+1)(k+ + n+1)(n+2) Let r2(n)=∑ t(1-t)(2-t) k(k+1)(k+2)(k+ (7) n=r2(n)+a1 n+1(n+1)(n+2) Define for n∈N,m≥2, rm(m)=∑ t(1-t)(2-t)…(m-t) k(k+1)(k+2)…(k+m)(k+t) dt, am t(1-t)…(m-1-t)dt.(9)

2 LI YINGYING From (∗) we obtain rn = X∞ k=n+1 Z 1 0 t k(k + t) dt = X∞ k=n+1 Z 1 0 t µ 1 k(k + t) − 1 k(k + 1)¶ dt + Z 1 0 tdt X∞ k=n+1 1 k(k + 1) = X∞ k=n+1 Z 1 0 t(1 − t) k(k + 1)(k + t) dt + 1 n + 1 Z 1 0 tdt. Write r1(n) = X∞ k=n+1 Z 1 0 t(1 − t) k(k + 1)(k + t) dt, a1 = Z 1 0 tdt. (5) Then rn = r1(n) + a1 n + 1 . (6) Moreover, r1(n) = X∞ k=n+1 Z 1 0 t(1 − t) k(k + 1)(k + t) dt = X∞ k=n+1 Z 1 0 t(1 − t) µ 1 k(k + 1)(k + t) − 1 k(k + 1)(k + 2)¶ dt + X∞ k=n+1 Z 1 0 t(1 − t) k(k + 1)(k + 2)dt = X∞ k=n+1 Z 1 0 t(1 − t)(2 − t) k(k + 1)(k + 2)(k + t) dt + X∞ k=n+1 1 2 Z 1 0 t(1 − t) dt µ 1 k(k + 1) − 1 (k + 1)(k + 2)¶ = X∞ k=n+1 Z 1 0 t(1 − t)(2 − t) k(k + 1)(k + 2)(k + t) dt + 1 2 Z 1 0 t(1 − t) dt 1 (n + 1)(n + 2). Let r2(n) = X∞ k=n+1 Z 1 0 t(1 − t)(2 − t) k(k + 1)(k + 2)(k + t) dt, a2 = 1 2 Z 1 0 t(1 − t)dt. (7) Then rn = r2(n) + a1 n + 1 + a2 (n + 1)(n + 2). (8) Define for m ∈ N, m ≥ 2, rm(n) = X∞ k=n+1 Z 1 0 t(1 − t)(2 − t)· · ·(m − t) k(k + 1)(k + 2)· · ·(k + m)(k + t) dt, am = 1 m Z 1 0 t(1 − t)· · ·(m − 1 − t) dt. (9)

ON EULER'S CONSTANT -CALCULATING SUMS BY INTEGRALS Then by induction we get (+1(m+2…(+6+m( From(5) and() we have k(k+1)2 (m+1)am+1(k-1>mm+1)am1y(k-2 k+1)(k+m)! n+2 + m Since k=7+1 (h +m) (n-1)! (k-1)k…(k+m (k+1 m)(m+1)(n+m)! we have > +2)(n+ On the other hand, by ( 9) we have obviously that for m22 2)!≤ (15) Then we conclude for m> 2 6(n+2)m(m+1)(m <Tm(n)< 6n(m+1)(m+2) where Taking(11) into account we see that(16)is also valid for m= 1. From(16)we get lim rm(n)=0 Then we have established the following theorem

ON EULER’S CONSTANT —CALCULATING SUMS BY INTEGRALS 3 Then by induction we get rn = Xm k=1 ak (n + 1)(n + 2)· · ·(n + k) + rm(n). (10) From (5) and (7) we have 2a2 X∞ k=n+1 1 k(k + 1)2 X∞ k=n+1 (m + 1)am+1(k − 1)! (k + 1)(k + m)! > n(m + 1)am+1 n + 2 X∞ k=n+1 (k − 2)! (k + m)!. Since X∞ k=n+1 (k − 2)! (k + m)! = X∞ k=n+1 1 (k − 1)k · · ·(k + m) = X∞ k=n+1 1 (m + 1) ³ 1 (k − 1)k · · ·(k + m − 1) − 1 k(k + 1)· · ·(k + m) ´ = (n − 1)! (m + 1)(n + m)!, we have rm(n) > am+1n! (n + 2)(n + m)!. (14) On the other hand, by (9) we have obviously that for m ≥ 2 1 6m (m − 2)! ≤ am ≤ 1 6m (m − 1)!. (15) Then we conclude for m ≥ 2 1 6(n + 2)m(m + 1)¡m+n m ¢ < rm(n) < 1 6n(m + 1)¡m+n m ¢ (16) where µ m + n m ¶ = m!n! (m + n)!. Taking (11) into account we see that (16) is also valid for m = 1. From (16) we get lim m→∞ rm(n) = 0. Then we have established the following theorem

LI YINGYING Theorem. Let Dn= k=1 k-log(n+1)(nEN)and y=limn-oo Dn be Euler constant. Then Tn: =?-Dn= 1(n+1)…(m+k) k t(1-t)…(k-1-t)dt(k>1) Furthermore 6(n+2)m(m+1)("m 之(+1)…(m+6)561(m+1)(m+7) For the numbers ak the referee provides the following table 1375 120a5=2046=504a7=168 He also points out that ak can be expressed in terms of Stirling numbers of the first kind s(k,j)as (-1)+y(k +. Our proof is completely an elementary calculation applying integral to estimate the sums. The method be ally applied to othe Acknowledgement The author is grateful to Professor Wang who enlightened her to write down this paper; also, she thanks the referee for valuable suggestions and for correcting printing mistakes refereNCeS 1. R. M. Young, Euler's Constant Math. Gazette 75 No 472(1991), 187-190 2. D W. De Temple, A quicker convergence to Euler's Constant, The Amer. Math. Monthly 100(1993),468-470 3. Rippon P L, Convergence with pictures, The Amer. Math. Monthly 93(1986), 476-478 4. D.W. De Temple and S.H. Wang, Half integer approximations for the partial sums of the harmonic series, J. Math lysis and Applic. 160(1991), 149-156 Department of Mathematics, Beijing Normal University, Beijing 100875, P.R. China ying-dd-Ii@263.net

4 LI YINGYING Theorem. Let Dn = Pn k=1 1 k − log(n + 1) (n ∈ N) and γ = limn→∞ Dn be Euler constant. Then rn := γ − Dn = X∞ k=1 ak (n + 1)· · ·(n + k) , where a1 = 1 2 , ak = 1 k Z 1 0 t(1 − t)· · ·(k − 1 − t) dt (k > 1). Furthermore 1 6(n + 2)m(m + 1)¡m+n m ¢ < rn − Xm k=1 ak (n + 1)· · ·(n + k) < 1 6n(m + 1)¡m+n m ¢ . For the numbers ak the referee provides the following table a1 = 1 2 , a2 = 1 12 , a3 = 1 12 , a4 = 19 120 , a5 = 9 20 , a6 = 863 504 , a7 = 1375 168 , a8 = 33953 720 . He also points out that ak can be expressed in terms of Stirling numbers of the first kind s(k, j) as ak = (−1)k+1 k X k j=1 s(k, j) j + 1 . Our proof is completely an elementary calculation applying integral to estimate the sums. The method may be generally applied to other cases. Acknowledgement The author is grateful to Professor Wang who enlightened her to write down this paper ; also , she thanks the referee for valuable suggestions and for correcting printing mistakes. References 1. R.M.Young, Euler’s Constant, Math. Gazette 75 No.472 (1991), 187–190. 2. D.W. DeTemple, A quicker convergence to Euler’s Constant, The Amer. Math. Monthly 100 (1993), 468–470. 3. Rippon P L, Convergence with pictures, The Amer.Math.Monthly 93 (1986), 476–478. 4. D.W. DeTemple and S.H. Wang, Half integer approximations for the partial sums of the harmonic series, J. Math. Analysis and Applic. 160 (1991), 149–156. Department of Mathematics, Beijing Normal University, Beijing 100875, P.R. China ying-dd-li@263.net