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Algebra Difficulty 5.3 AIME, harder Prove it Romania

Let (an)n1(a_n)_{n \ge 1} be an increasing bounded sequence of real numbers. Evaluate
limn(2ana1a2)(2ana2a3)(2anan2an1)(2anan1a1). \lim_{n \to \infty} (2a_n - a_1 - a_2)(2a_n - a_2 - a_3) \cdots (2a_n - a_{n-2} - a_{n-1})(2a_n - a_{n-1} - a_1).

Solution

The limit is equal to 00.
Let xn=(2ana1a2)(2ana2a3)(2anan2an1)(2anan1a1)x_n = (2a_n - a_1 - a_2)(2a_n - a_2 - a_3) \cdots (2a_n - a_{n-2} - a_{n-1})(2a_n - a_{n-1} - a_1). The sequence (an)n1(a_n)_{n \ge 1} is convergent; let L=limnanL = \lim_{n \to \infty} a_n. Since anLa_n \le L for all n1n \ge 1 we have 2anakak+12Lakak+12(Lak)2a_n - a_k - a_{k+1} \le 2L - a_k - a_{k+1} \le 2(L - a_k), k=1,2,,n2k = 1, 2, \dots, n-2, so 0xn2n1(La1)(La2)(Lan2)(La1)=yn0 \le x_n \le 2^{n-1}(L - a_1)(L - a_2) \cdots (L - a_{n-2})(L - a_1) = y_n.
Suppose yn>0y_n > 0, otherwise yn=0y_n = 0 and then xn=0x_n = 0. Since yn+1yn=2(Lan)0\frac{y_{n+1}}{y_n} = 2(L - a_n) \to 0, we infer that limnyn=0\lim_{n \to \infty} y_n = 0 and finally limnxn=0\lim_{n \to \infty} x_n = 0.

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