Do either (1) or (2)
(1) Prove that ∑2∞ cos(ln ln n) / ln n diverges.
(2) Let k, a, b, c be real numbers such that a, k > 0 and b2 < ac. Show that ∫U (k + ax2 + 2bxy + cy2)-2 dx dy = π/( k √(ac - b2) ), where U is the entire plane.
Solution
(1) We have ln n2 = 2 ln n, so for any m in the range n, n+1, n+2, ... , n2 we have 1/ln m > 1/2 1/ln n. Also ln ln n2 = ln ln n + ln 2 < ln ln n + π/3. So given any sufficiently large integer N, take x > N so that ln ln x = a multiple of 2 π, then we can find at least N2/2 consecutive integers n in the range x to x2such that (A) 1/ln n > 1/(2 ln N) and (B) ln ln n lies between 2k π and 2k π + π/3 (for some integer k) and hence so that cos ln ln n > 1/2. Hence for each such integer (cos ln ln n)/ln n > 1/(4 ln N). So ∑ (cos ln ln n)/ln n over all these terms is at least N2/(8 ln N) which can be made arbitrarily large. Hence the sequence does not converge.
(2) ax2 + 2bxy + cy2 = h is the equation of an ellipse. If we rotate the axes to line up with its axes we get new coordinates X, Y so that it becomes AX2 + BY2 = h with AB = ac - b2. Thus the integral is transformed to ∫U (k + AX2 + BY2)-2 dX dY. Take u = X√A, v = Y√B and the integral becomes 1/√(ac - b2) ∫U (k + u2 + v2)-2 du dv. Now take u = r cos θ, v = r sin θ and we get 1/√(ac - b2) ∫U (k + u2 + v2)-2 du dv = 1/√(ac - b2) ∫U (k + r2)-2 r dr dθ = 2 π /√(ac - b2) ∫ (k + r2)-2 r dr = 2 π /(k √(ac - b2) ) ∫ (1 + s2)-2 s ds = π /(k √(ac - b2) ) (-1/(1 + s2)|0inf = 2 π /(k √(ac - b2) ).
Thanks to Dave Rusin for a similar solution.
© John Scholes
jscholes@kalva.demon.co.uk
11 Mar 2002