证明了{n(16n^2+4n+3)/16n^2-4~n+3^(1/2) integral from 0 to π/2 sin^nxdx}为严格单调增加数列,且极限为π/2^(1/2),因而得π(16n^2+36n+23)/2(n+1)(16n^2+28n+15)^(1/2)<integral from 0 to π/2 sin^nxdx<π(16n^2-4n+3)/2n(...证明了{n(16n^2+4n+3)/16n^2-4~n+3^(1/2) integral from 0 to π/2 sin^nxdx}为严格单调增加数列,且极限为π/2^(1/2),因而得π(16n^2+36n+23)/2(n+1)(16n^2+28n+15)^(1/2)<integral from 0 to π/2 sin^nxdx<π(16n^2-4n+3)/2n(16n^2+4n+3)^(1/2).展开更多
文摘证明了{n(16n^2+4n+3)/16n^2-4~n+3^(1/2) integral from 0 to π/2 sin^nxdx}为严格单调增加数列,且极限为π/2^(1/2),因而得π(16n^2+36n+23)/2(n+1)(16n^2+28n+15)^(1/2)<integral from 0 to π/2 sin^nxdx<π(16n^2-4n+3)/2n(16n^2+4n+3)^(1/2).