n → ∞ lim ( n 2 + 1 2 n + n 2 + 2 2 n + n 2 + 3 2 n + ⋯ + n 2 + n 2 n ) = ?
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Thank you again for the elegant and clear solution :)
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Nice to know that you like it. Thanks for the problem.
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Relevant wiki: Riemann Sums
L = n → ∞ lim k = 1 ∑ n n 2 + k 2 n = n → ∞ lim k = 1 ∑ n n + n k 2 1 = n → ∞ lim n 1 k = 1 ∑ n 1 + ( n k ) 2 1 = ∫ a b 1 + x 2 1 d x = ∫ 0 1 1 + x 2 1 d x = tan − 1 x ∣ ∣ ∣ ∣ 0 1 = 4 π Divide up and down by n By Riemann sums a = n → ∞ lim n 1 = 0 , b = n → ∞ lim n n = 1