# 数列极限原题如下an=ln(13)+ln(24)+ln(35)+...+ln(nn+2)a_n=ln(\frac{1}{3})+ln(\frac{2}{4})+ln(\frac{3}{5})+...+ln(\frac{n}{n+2})an=ln(31)+ln(42)+ln(53)+...+ln(n+2n) .有上界无下界有下界无上界无上界无下界有上界有下界将 lnxlnxlnx 除法改写为加减法,得到 ln1+ln2−ln(n−1)−ln(n−2)ln1+ln2-ln(n-1)-ln(n-2)ln1+ln2−ln(n−1)−ln(n−2)ln1+ln2−ln(n−1)−ln(n−2)=ln(2n2+3n+2)≤0ln1+ln2-ln(n-1)-ln(n-2) = \ln(\frac{2}{n^2+3n+2}) \le 0ln1+ln2−ln(n−1)−ln(n−2)=ln(n2+3n+22)≤0由于 n2+3n+2>2emn^2+3n+2>\frac{2}{e^m}n2+3n+2>em2, 所以 ln(2n2+3n+2)≤m\ln(\frac{2}{n^2+3n+2}) \le mln(n2+3n+22)≤m, 故 ana_nan 无下界.数列 xnx_nxn, yny_nyn 满足 xn+1≥2Xnπx_{n+1} \ge \frac{2X_n}{\pi}xn+1≥π2Xn, 0<yn+1≤2Ynπ0 < y_{n+1} \le \frac{2Y_n}{\pi}0<yn+1≤π2Yn, n=1,2,3,...,nn = 1,2,3,...,nn=1,2,3,...,n. 已知x1=y1=12x_1=y_1=\frac{1}{2}x1=y1=21, 当n→∞n\to \inftyn→∞ .yny_nyn 与xnx_nxn 为同阶无穷小yn=o(xn)y_n = o(x_n)yn=o(xn) 但不一定=o(xn2)=o(x^2_{n})=o(xn2)xn=o(yn)x_n = o(y_n)xn=o(yn) 但不一定=o(yn2)=o(y^2_{n})=o(yn2)yn=o(xn2)y_n = o(x_n^2)yn=o(xn2) 但不一定=o(xn3)=o(x^3_{n})=o(xn3)大道五十,天衍四十九,人遁其一! 考研,数学