IMO 1970 SL 10

Origin: SWE

Problem

Let 1 = a0 \leqa1 \leqa2 \leq\cdot \cdot \cdot \leqan \leq\cdot \cdot \cdot be a sequence of real numbers. Consider the sequence b1, b2, . . . defined by: bn = n  k=1  1 −ak−1 ak  \sqrtak . Prove that: (a) For all natural numbers n, 0 \leqbn < 2. (b) Given an arbitrary 0 \leqb < 2, there is a sequence a0, a1, . . . , an, . . . of the above type such that bn > b is true for infinitely many natural numbers n.

Solution

(a) Since an−1 < an, we have  1 −ak−1 ak  \sqrtak = ak −ak−1 a3/2 k \leq2(\sqrtak −\sqrtak−1)\sqrtak ak\sqrtak−1 = 2  \sqrtak−1 − \sqrtak  . Summing up all these inequalities for k = 1, 2, . . . , n we obtain bn \leq2  1 \sqrta0 − \sqrtan  < 2. (b) Choose a real number q > 1, and let ak = qk, k = 1, 2, . . . . Then (1 −ak−1/ak) /\sqrtak = (1 −1/q) /qk/2, and consequently bn =  1 −1 q  n  k=1 qk/2 = \sqrtq + 1 q  1 − qn/2  . Since (\sqrtq + 1)/q can be arbitrarily close to 2, one can set q such that (\sqrtq + 1)/q > b. Then bn \geqb for all sufficiently large n. Second solution. (a) Note that bn = n  k=1  1 −ak−1 ak  \sqrtak

n  k=1 (ak −ak−1) \cdot a3/2 k ;

hence bn represents exactly the lower Darboux sum for the function f(x) = x−3/2 on the interval [a0, an]. Then bn \leq = an a0 x−3/2dx < = +\infty x−3/2dx = 2. (b) For each b < 2 there exists a number \alpha > 1 such that = \alpha 1 x−3/2dx > b + (2 −b)/2. Now, by Darboux’s theorem, there exists an array 1 = a0 \leqa1 \leq\cdot \cdot \cdot \leqan = \alpha such that the corresponding Darboux sums are arbitrarily close to the value of the integral. In particular, there is an array a0, . . . , an with bn > b.