Exercises on Sequences

1. Let A > 0. The sequence defined by

   a_n+1 = (1/2)(a_n + A/a_n)  (E.1)

   with a₁ ≥ 0. This sequence is used to calculate approximations to √A, and it also known as the Babylonian Algorithm for computing square roots. Determine whether the sequences converges and eventualy its limit

Solutions

1. Recalling the inequality relating geometric and arithmetic mean of two numbers a,b > 0

   $$\sqrt{ab}\le \frac{a+b}{2}$$

   In this inequality, put a = a_n and b = A/a_n to deduce

   √A ≤ (1/2) (a_n + A/a_n) = a_{n + 1}

   So the sequence is bounded below by √A at least starting from n = 2. To prove the convergence it remains to show it is decreasing, which is the same as showing that a_n+1 − a_n < 0. In fact, for n ≥ 2 we have a_n² > A, so that

   a_n+1 = (1/2) (a_n + A/a_n) ≤ (a_n + a_n)/2 = a_n

   Hence by theorem Monotone Sequence Theorem 2.2.5., the sequence a_n converges to some number L ≥ √A. Taking the limit on both sides of E.1, we deduce that L satisfies L = (1/2)(L + A/L) and so L = √A.  ■

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