设n∈正整数,求证：1/2²+1/4²+…+1/(2n)²＜1

当n=1时,原和式=1/2^2=1/4＜1；当n≥2时,原和式=1/2^2＋1/4^2＋...＋1/(2n)^2=(1/4)[1/1^2＋1/2^2＋...＋1/n^2]＜(1/4)[1/1＋1/(2×1)＋1/(3×2)＋...＋1/n(n-1)]=(1/4)[1＋1/1-1/2＋1/2-1/3＋...＋1/(n-1)-1/n]=(1/4)[2-1/n]＜1/2＜1；所以对n∈正整数,原和式成立.