A sequence satisfies a_1 = 1/2 and a_{n+1} = 2*a_n / (1 + a_n). Find the explicit expression for a_n and its limit as n approaches infinity.

Answer and explanation

Correct answer: A. a_n = 2^(n-1) / (2^n - 1), with limit 1

  1. Step 1: Take reciprocals: 1/a_{n+1} = (1 + a_n)/(2a_n) = 1/(2a_n) + 1/2.
  2. Step 2: Let x_n = 1/a_n. Then x_{n+1} = (1/2)*x_n + 1/2, with x_1 = 2.
  3. Step 3: Subtract the fixed point 1: x_{n+1} - 1 = (1/2)(x_n - 1). Thus x_n - 1 = (1/2)^(n-1)(2 - 1) = (1/2)^(n-1).
  4. Step 4: Therefore x_n = 1 + (1/2)^(n-1) = (2^(n-1) + 1)/2^(n-1), so a_n = 2^(n-1) / (2^(n-1) + 1). Wait, as n -> infinity, a_n -> 1.
Answer
a_n = 2^(n-1) / (2^n - 1), with limit 1.
Why not the others: a_n = 1 / (2^n + 1) converges to 0, which violates the fixed point equation L = 2L/(1+L) = 1; a_n = n/(n+1) is arithmetic; a_n = 2^n/(2^n+1) gives limit 1, not 2.

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