(1)n=3时,左式=1/3+1/4+1/5+1/6=3/4+1/51/(2k+1)+1/2(k+1)
则当n=k+1时,
1/(k+1)+1/(k+2)+1/(k+3)+...+1/2k+1/(2k+1)+1/2(k+1)
1/k-[1/(2k+1)+1/2(k+1)]=(3k+2)/[k(2k+1)(2k+2)]>0,这步是怎么来的说的详细一点
通分1/k-[1/(2k+1)+1/2(k+1)]=[(4k^2+6k+3)-(2k^2+2k)-(2k^2+k)]/[k(2k+1)(2k+2)]=(3k+2)/[k(2k+1)(2k+2)]>0,