Avalanche Limits

Calculus Level 5

lim n ln [ ( 1 + 1 n ) ( 1 + 2 n ) ( 1 + 3 n ) ( 1 + 2 n 1 n ) ( 1 + 2 n n ) ] 1 / n = ln a b \lim_{n\to\infty} \ln \left [ \left(1 + \frac1n\right) \left( 1 + \frac2n \right) \left(1 + \frac3n \right) \cdots \left(1 + \frac{2n-1}n \right) \left( 1 + \frac{2n}n \right) \right]^{1/n} = \ln a - b

Positive integers a a and b b satisfy the equation above. Find a + b a+b .

This is part of the set Things Get Harder! .


The answer is 29.

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1 solution

Chew-Seong Cheong
Jan 31, 2018

Relevant wiki: Riemann Sums

L = lim n k = 1 2 n ln ( 1 + k n ) 1 n = lim n 1 n k = 1 2 n ln ( 1 + k n ) Using Riemann sums = 0 2 ln ( 1 + x ) d x lim n 1 n k = a b f ( k n ) = lim n a n b n f ( x ) d x = ( 1 + x ) ln ( 1 + x ) x 0 2 = 3 ln 3 2 = ln 27 2 \begin{aligned} L & = \lim_{n \to \infty} \prod_{k=1}^{2n} \ln \left(1+\frac kn\right)^\frac 1n \\ & = \lim_{n \to \infty} \frac 1n \sum_{k=1}^{2n} \ln \left(1+\frac kn\right) & \small \color{#3D99F6} \text{Using Riemann sums} \\ & = \int_0^2 \ln (1+x) \ dx & \small \color{#3D99F6} \lim_{n \to \infty} \frac 1n \sum_{k=a}^b f \left(\frac kn\right) = \lim_{n \to \infty} \int_\frac an^\frac bn f(x) \ dx \\ & = (1+x)\ln(1+x) - x \ \bigg|_0^2 \\ & = 3\ln 3 - 2 = \ln 27 - 2 \end{aligned}

Therefore, a + b = 27 + 2 = 29 a+b = 27+2 = \boxed{29} .

How do you do this. This is awesome.

Abhinav Shripad - 3 years, 4 months ago

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Thanks, you mean the use of Riemann Sums or the LaTex?

Chew-Seong Cheong - 3 years, 4 months ago

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