zoj 1007 Numerical Summation ...

#include <stdio.h>int main() {double sum = 0;double x;int k; for(x = 0.000; x <= 2.000; x += 0.001) {sum = 0;for(k= 1; k < 10000; k++) {sum += (1.0 - x) / (k * (k + 1) * (k + x));}sum = sum + (1 - x) / (2 * 10000 * 10000) + 1.0;printf("%5.3f %16.12fn", x, sum); }return 0;}