n → ∞ lim ( n 2 + 1 n + n 2 + 4 n + n 2 + 9 n + ⋯ + n 2 + n 2 n )
Find the value of the closed form of the above limit.
Give your answer to 3 decimal places.
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Relevant wiki: Riemann Sums
L = n → ∞ lim ( n 2 + 1 n + n 2 + 4 n + n 2 + 9 n + ⋯ + n 2 + n 2 n ) = n → ∞ lim k = 1 ∑ n n 2 + k 2 n = n → ∞ lim k = 1 ∑ n 1 + n 2 k 2 n 1 = n → ∞ lim n 1 k = 1 ∑ n 1 + n 2 k 2 1 = ∫ a = 0 b = 1 1 + x 2 1 d x = tan − 1 x ∣ ∣ ∣ ∣ 0 1 = 4 π ≈ 0 . 7 8 5 Divide up and down with n 2 By Riemann sums a = n → ∞ lim n k = 0 , b = n → ∞ lim n n = 1