n → ∞ lim ln [ ( 1 + n 1 ) ( 1 + n 2 ) ( 1 + n 3 ) ⋯ ( 1 + n 2 n − 1 ) ( 1 + n 2 n ) ] 1 / n = ln a − b
Positive integers a and b satisfy the equation above. Find a + b .
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How do you do this. This is awesome.
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Thanks, you mean the use of Riemann Sums or the LaTex?
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Relevant wiki: Riemann Sums
L = n → ∞ lim k = 1 ∏ 2 n ln ( 1 + n k ) n 1 = n → ∞ lim n 1 k = 1 ∑ 2 n ln ( 1 + n k ) = ∫ 0 2 ln ( 1 + x ) d x = ( 1 + x ) ln ( 1 + x ) − x ∣ ∣ ∣ ∣ 0 2 = 3 ln 3 − 2 = ln 2 7 − 2 Using Riemann sums n → ∞ lim n 1 k = a ∑ b f ( n k ) = n → ∞ lim ∫ n a n b f ( x ) d x
Therefore, a + b = 2 7 + 2 = 2 9 .