WHAT ARE THE LIMITS? SOLUTION TO PROBLEM 11811 OF THE AMERICAN MATHEMATICAL MONTHLY
OMRAN KOUBA Abstract. We determine the limit points of the sequence (xn )n≥1 given by xn =
ab11 · · · abnn a1 b1 + · · · + an bn b1 + · · · + bn
b1 +···+bn
where (an )n≥1 and (bn )n≥1 are arbitrary positive sequences.
Problem 11811 [1]. Proposed by Vazgen Mikayelyan, Yerevan State University, Yerevan, Armenia. Let hai and hbi be infinite sequences of positive numbers. Let hxi be the infinite sequence given for n ≥ 1 by xn =
ab11 · · · abnn a1 b 1 + · · · + an b n b1 + · · · + bn
b1 +···+bn
(a) Prove that limn→∞ xn exists. (b) Find the set of all c that can occur as that limit, for suitably chosen hai and hbi. Solution [2]: (a) We first claim that hxi is monotonic nonincreasing. With notation: !1/Λn n n Y bn 1 X Λn = b1 + · · · + bn , λn = , An = b k ak . , Gn = abkk Λn Λ n k=1 k=1 we have xn =
Gn An
Λn .
(1)
λ
n+1 n+1 and An+1 = (1 − λn+1 )An + λn+1 an+1 , hence Clearly we have Gn+1 = G1−λ an+1 n
λ
n+1 n+1 G1−λ an+1 Gn+1 n = = An+1 (1 − λn+1 )An + λn+1 an+1
1
Gn An
1−λn+1
λ
n+1 n+1 A1−λ an+1 n (1 − λn+1 )An + λn+1 an+1
2
OMRAN KOUBA
or equivalently xn+1 = xn
λ
n+1 n+1 an+1 A1−λ n (1 − λn+1 )An + λn+1 an+1
!Λn+1 (2)
By the arithmetic mean-geometric mean inequality, we have A1−λ aλ ≤ (1 − λ)A + λa for positive a, A and λ ∈ (0, 1), so from (2) we conclude that xn+1 ≤ xn (with equality if and only if an+1 = An .) Moreover, note that x1 = 1 and xn > 0 for all n ≥ 1. Therefore, hxi converges and the limit belongs to [0, 1]. (b) It suffices to prove the following: if c ∈ [0, 1], then there exist sequences hai and hbi such that hxi converges to c. 1 If c = 0, then let an = n(n+1) and bn = 1 for all n, so (n + 1)n −→ 0. n! · (n + 1)! n→∞ If c = 1, then let an = 1 and bn = 1 for all n, so xn = 1 for all n ≥ 1 and limn→∞ xn = 1. If 0 < c < 1 we consider the sequences hbi and hai defined by bn = 1 for n ≥ 1 and √ √ 1+ 1−c 1− 1−c 1 √ √ a1 = , a2 = , and an = √ for n ≥ 3. c c c It is straightforward to check that hxi is given by x1 = 1 and xn = c for n ≥ 2. So limn→∞ xn = c in this case. xn =
References [1] Vazgen Mikayelyan, Proposed Problem 11811, The American Mathematical Monthly, Vol. 122, No. 1, (2015), p.75. [2] Omran Kouba, What the limits are?, The American Mathematical Monthly, Vol. 123, No. 8, (2016), p. 837. Doi:10.4169/amer.math.monthly.123.8.831 Department of Mathematics, Higher Institute for Applied Sciences and Technology, P.O. Box 31983, Damascus, Syria. E-mail address: omran
[email protected] 