Show that limn→∞ 1/n ∑1n ( [2n/i] - 2[n/i] ) = ln a - b for some positive integers a and b.
Solution
Answer: ln 4 - 1.
The expression inside the limit is a partial sum for ∫01 [2/x] - [1/x] dx (taking the points x = 1/n, 2/n, ... , 1). So the limit is ∫01 [2/x] - [1/x] dx.
Evidently the integrand is non-zero on the intervals (1/(n+1), 1/(n + 1/2) ], for if x belongs to such an interval then n + 1/2 <= x < n + 1, so [1/x] = n, whilst 2n + 1 ≤ x ≤ 2n + 2, so [2/x] = 2n+1. A similar argument shows it is zero on the complementary intervals (1/(n + 1/2), 1/n]. Thus the integral evaluates to (1/(1 + 1/2) - 1/2) + (1/(2 + 1/2) - 1/2 + ... = 2(1/3 - 1/4 + 1/5 - ... ) = 2(1 - 1/2 + 1/3 - 1/4 + ... ) - 1 = 2 ln 2 - 1 = ln 4 - 1.
© John Scholes
jscholes@kalva.demon.co.uk
23 Jan 2001