lim n → ∞ 1 1 + n 5 + 2 4 2 5 + n 5 + 3 4 3 5 + n 5 + … + n 4 n 5 + n 5 =

Question

lim n → ∞ 1 1 + n 5 + 2 4 2 5 + n 5 + 3 4 3 5 + n 5 + … + n 4 n 5 + n 5 =, 1 5 log 3, 1 3 log 5, 1 2 log 5, log 2 5

MARKS App · Accepted Answer

lim n → ∞ 1 1 + n 5 + 2 4 2 5 + n 5 + 3 4 3 5 + n 5 + … + n 4 n 5 + n 5 = lim n → ∞ ∑ r = 1 n r 4 r 5 + n 5 = lim n → ∞ 1 n ∑ r = 1 n r n 4 r n 5 + 1 = ∫ 0 1 x 4 1 + x 5 d x = 1 5 ln 1 + x 5 0 1 = 1 5 ln 2 = ln 2 5

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