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Extra info for Problems in Real Analysis: Advanced Calculus on the Real Axis
An−1 + a +x n n ⎟ ⎟. ⎠ By (ii), the function gk−2 is increasing, so its maximum is achieved for ak = 1. 1+ 1+x n Fk ∑ Fk+1 . k=1 This concludes our proof. A simple linear recurrence generates a sequence of perfect squares, as shown below. 8. A sequence of integers (an )n≥1 is given by the conditions a1 = 1, a2 = 12, a3 = 20, and an+3 = 2an+2 + 2an+1 − an for every n ≥ 1. Prove that for every positive integer n, the number 1 + 4anan+1 is a perfect square. Problem M1174∗, Kvant Solution. Define the sequence (bn )n≥1 by bn = an+2 − an+1 − an , for any n ≥ 1.
Y|2 2 An interesting convergent rearrangement is possible, provided a sequence of positive real numbers has a finite nonzero accumulation point. 18. Let (an )n≥1 be a sequence of positive real numbers that has a finite nonzero accumulation point. Prove that (an )n≥1 can be arranged in a sequence 1/n (xn )n≥1 so that (xn )n≥1 is convergent. Solution. Let a ∈ (0, ∞) be an accumulation point of (an )n≥1 and choose α > 0 such that α −1 < a < α . Let x1 denote the an of smallest subscript that lies in the interval (α −1 , α ).
Prove that f (x) tends to zero as x tends to infinity. A complete proof of this result will be given in Chapter 10. 4 Qualitative Results A mathematician who is not also a poet will never be a complete mathematician. , Carleman’s inequality in the next section). We illustrate here this result to deduce an interesting inequality involving positive numbers. We point out that a very important international prize was created by the Norwegian Academy of Science and Letters in order to celebrate the Abel centenary in 2002.