137_PS1_Sequences_SOLUTIONS.pdf - Problem Sets u2013...

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Problem Sets – Sequences Explicitly-defined sequences 1. Determine whether the following sequences converge, and if they do, find the limit. a) a n = 3 n n + 2 , b) a n = 3 + 5 n 2 1 + n , c) a n = 1 + 2 n n SOLUTIONS: a) Divide the numerator and denominator by n , a n = 3 n n + 2 = 3 1 + 2 / n The numerator and the denominator are convergent sequences, and so the limit of the original sequence is a quotient of their individual limits, lim n →∞ 3 n n + 2 = lim n →∞ 3 lim n →∞ 1 + 2 / n = 3 1 = 3 b) Divide the numerator and the denominator by n 2 , a n = 3 + 5 n 2 1 + n = 3 /n 2 + 5 1 /n 2 + 1 /n Again, the numerator and the denominator are convergent sequences, BUT the limit of the denominator is zero, 1 /n 2 + 1 /n 0 and n → ∞ . Consequently, the sequence diverges ( i.e. does not converge). c) To apply the limit rules, first make the change of variable n/ 2 7→ k (and so n 7→ 2 k ), then the original sequence is transformed to a n = 1 + 2 n n 7→ b k = 1 + 1 k 2 k = " 1 + 1 k k # 2 The limit of the sequence inside the square brackets is the natural exponential base e = 2 . 718281828 . . . . The full limit is then, lim n →∞ a n = lim k →∞ b k = lim k →∞ " 1 + 1 k k # 2 = " lim k →∞ 1 + 1 k k # 2 = e 2 = 7 . 389056099 . . . Implicitly-defined sequences 2. Determine whether the following sequence converges and, if so, determine the limit. 2 , q 2 2 , r 2 q 2 2 , s 2 r 2 q 2 2 , . . . SOLUTION: Usually implicitly-defined sequences are difficult to analyze directly, but the initial term of this sequence makes our work much easier. First, from the list of terms, the rule generating the sequence seems to be: a n +1 = 2 a n ( n 0) , with a 0 = 2 Using the exponential form for the square-root, a n +1 = (2 a n ) 1 / 2 ( n 0) , with a 0 = 2 1 / 2
Writing out a few terms, a 0 = 2 1 / 2 a 1 = 2 × 2 1 / 2 1 / 2 = 2 3 / 2 1 / 2 = 2 3 / 4 a 2 = 2 × 2 3 / 4 1 / 2 = 2 7 / 4 1 / 2 = 2 7 / 8 a 3 = 2 15 / 16 . . . a n = 2 2 ( n +1) - 1 2 ( n +1) = 2 1 - 1 2 ( n +1) Each term is 2 raised to a fraction with a power of two in the denominator, and one-minus-a-power-of-two in the numerator. As n → ∞ , the exponent approaches 1, and so lim n →∞ a n = 2 1 = 2 3. An implicitly-defined sequence is given by a n +1 = 2 + a n ( n 1), with initial term a 1 = 2. a) Show that the sequence is increasing and bounded above by 3. SOLUTION: The claim is that 3 a n +1 a n It is certainly true for the first two terms: a 1 = 2 = 1 . 41421 . . . and a 2 = p 2 + 2 = 1 . 84776 . . . , so 3 a 2 a 1 X For the ‘induction’ part of the proof, given that 3 a n +1 a n , we must show that 3 a n +2 a n +1 is also true ( i.e. that the first statement implies the second). Starting from what we are given, 3 a n +1 a n we want to convert a n +1 a n +2 and a n a n +1 . To that end, add 2 to each term in the inequality, 2 + 3 2 + a n +1 2 + a n This step does not change the logic of the inequalities. The square-root function is a strictly-increasing function, and so it also won’t change the logic of the inequalities. Taking the square-root of each term, 2 + 3 p 2 + a n +1 2 + a n You will recognize the last two terms as a n +2 = 2 + a n +1 and a n +1 = 2 + a n so we have 2 + 3 a n +2 a n +1 But 2 + 3 = 5 < 3, so 3 2 + 3 a n +2 a n +1 X which means we have proved the claim by induction.

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