PROPOSED SOLUTION TO PROBLEMA 141 OVIDIU FURDUI Sea n i un numero combinatorio, n, i ∈ N n ≥ i. Demostrar entonces que n n n ··· 1 2 n 1 2 y e 2 n son dos infinitos asintoticamente equivalentes, es decir, 1 lim n→∞ 2 e2n n n 1 2 ··· n n = 1. Solution. In what follows we are going to prove that 2 1 lim n→∞ e2n n n 1 2 ··· n n = ∞. We need the following Lemma. Lemma 0.1. Glaisher-Kinkelin constant. The following limit holds: A = lim n→∞ n 11 22 · · · nn , e−n2 /4 n2 /2+n/2+1/12 where A = 1.282712 is the Glaisher-Kinkelin constant. 1 Let xn = 2 e2n . )(n2 )···(nn) n 1 ( ln xn = We have n n k=1 k=1 X n! n2 n2 X − ln = − (n + 1) ln n! + 2 ln k!. 2 k!(n − k)! 2 Date: May 7, 2007. 1 2 OVIDIU FURDUI On the other hand, we note that n X ln k! = n ln 1 + (n − 1) ln 2 + · · · + 2 ln(n − 1) + ln n = k=1 n X (n + 1 − k) ln k k=1 = (n + 1) ln n! − n X k ln k. k=1 Thus, n ln xn = X n2 + (n + 1) ln n! − 2 k ln k 2 k=1 ! 2 n X n n 1 n2 n2 k ln k − + (n + 1) ln n! − 2 + + ln n + = 2 2 2 12 4 k=1 1 n2 1 = n2 + (n + 1) ln n! − n2 + n + ln n + ln n − 2An , − n2 + n + 6 2 6 where An = n X k ln k − k=1 Using Stirling’s formula n! ≈ √ 2πn n n , e n2 n 1 + + 2 2 12 ln n + n2 . 4 we get that 1 n + (n + 1) ln n! − n + n + 6 2 2 √ ln n ln n ≈ (n + 1) ln 2πn − n − . 6 It follows, based on Lemma 0.1, that √ ln n ln xn ≈ (n + 1) ln 2πn − n − − 2An → ∞ − 2 ln A = ∞, 6 and hence xn → ∞. Western Michigan University, Kalamazoo, MI E-mail address: ofurdui@yahoo.com, o0furdui@wmich.edu Revista Escolar de la Olimpíada Iberoamericana de Matemática http://www.campus-oei.org/oim/revistaoim/ Edita: