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The value of l i m n → ∞ 1 2 n + 1 2 n + 1 + 1 2 n + 2 + . . . . . + 1 4 n is equal to

Options

  1. Ae 2
  2. Bln ⁡ 2
  3. Cln ⁡ 4
  4. D3 ln ⁡ 2

Correct answer

B. ln ⁡ 2

Step-by-step solution

l i m n → ∞ ∑ r = 0 2 n 1 2 n + r = l i m n → ∞ 1 n ∑ r = 0 2 n 1 2 + r n = ∫ 0 2 1 2 + x d x = ln ⁡ x + 2 0 2 = ln ⁡ 4 - ln ⁡ 2 = ln ⁡ 2

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