JEEOlympiad
Let L = lim_{n->infinity} (1/n)·( (n+1)(n+2)...(2n) )^(1/n). Compute…
Question
Let L = lim_{n->infinity} (1/n)·( (n+1)(n+2)...(2n) )^(1/n). Compute the integer value of floor(1000·L).
✓ Verified answer: 1471checked by our engine — not a guess
Step-by-step solution
Take logarithms. ln( (1/n)·( prod_{k=1}^{n} (n+k) )^(1/n) ) = -ln n + (1/n)·sum_{k=1}^{n} ln(n+k) = (1/n)·sum_{k=1}^{n} ln( (n+k)/n ) = (1/n)·sum_{k=1}^{n} ln(1 + k/n).
This is a Riemann sum for integral_0^1 ln(1+t) dt = [ (1+t)ln(1+t) - (1+t) ]_0^1 = (2 ln 2 - 2) - (0 - 1) = 2 ln 2 - 1.
So ln L = 2 ln 2 - 1, giving L = e^(2 ln 2 - 1) = 4/e ≈ 1.4715177....
Then floor(1000·L) = floor(1471.5177...) = 1471.
Final answer1471
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