编写一程序段求s=1/2+1/6+1/12+…+1/（n（n-1））+…直到最后1项的值小于1e-6.

main(){

float s=0;

float flag=1;

float i=2;

while(flag>0.000001){

s= s + (float)1/(float)(i*(i-1));

flag=(float)1/(float)(i*(i-1));

i++;

}

printf(" %f",s);

}