IMO 1985 SL 6

Origin: POL

Problem

Let xn = 2 + 3 + . . . + n\sqrtn. Prove that xn+1 −xn < 1 n!, n = 2, 3, . . . . Alternatives

Solution

Let us set xn,i = i  i + i+1 i + 1 + \cdot \cdot \cdot + n\sqrtn, yn,i = xi−1 n+1,i + xi−2 n+1,ixn,i + \cdot \cdot \cdot + xi−1 n,i . In particular, xn,2 = xn and xn,i = 0 for i > n. We observe that for n > i > 2, xn+1,i −xn,i = xi n+1,i −xi n,i yn,i = xn+1,i+1 −xn,i+1 yn,i . Since yn,i > ixi−1 n,i \geqi1+(i−1)/i \geqi3/2 and xn+1,n+1 −xn,n+1 = n+1\sqrtn + 1, simple induction gives xn+1 −xn \leq n+1\sqrtn + 1 (n!)3/2 < 1 n! for n > 2. The inequality for n = 2 is directly verified.