solutions1

# Solutions1 - CM221A ANALYSIS I Solutions to Sheet 1 1 By the Archimedean Property(see course notes we have 1 < n for some n N Multiplying both

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CM221A ANALYSIS I Solutions to Sheet 1 1. By the Archimedean Property (see course notes), we have δ - 1 < n for some n N . Multiplying both parts of this inequality by n - 1 δ , we obtain n - 1 < δ . 2. Obviously, each a n is positive and, given this, 0 < a n +1 < a n . The sequence, being monotonically decreasing and bounded, converges to a limit which satisﬁes a = a/ ( a + 2) (by passing to the limit in the identity deﬁning a n +1 ). Hence a = 0. 3. lim n →∞ 2 n 3 + n 2 + n - n 5 n 3 + 2 n - 1 + 2 n 5 = lim n →∞ 2 n - 2 + n - 3 + n - 4 - 1 n - 2 + 2 n - 4 - n - 5 + 2 . Evaluating the limits in the numerator and denominator, we obtain - 1 2 . lim n →∞ 2 n + 3 3 n + 2 = lim n →∞ (2 / 3) n 1 + 3 × 2 - n 1 + 2 × 3 - n = lim n →∞ (2 / 3) n 1 + 3 lim n →∞ × 2 - n 1 + 2 lim n →∞ 3 - n = 0 . 4. Let a n = n k b - n . Then a n +1 /a n = b - 1 ± n +1 n ² k . Since n +1 n 1 as n → ∞ , we see that there is a number c < 1 such that a m +1 /a m < c for all suﬃciently large m . It follows that there exists a
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## This document was uploaded on 01/31/2011.

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