求1+1/2+1/3++1 include <stdio h> :include <stdio. h> int main(void) int main(void) i int i; int i, sum, double sum printf(Enter n: ); printf((“ Enter n:" scanf( %d",&n); scanf( %d",&n); sum = 0: sum〓0 for(i=1; i<=n; i++) for(i=1; i<=n; 1++) sum sum 10/i sum sum f printf((“sum=%fn”,sum) printf((“sum=%dn”,sum) return 0 return o#include <stdio.h> int main(void) { int i, sum; printf(“Enter n: "); scanf("%d", &n); sum = 0; for ( i = 1; i <= n; i++ ) sum = sum + i; printf( “sum = %d\n”, sum); return 0; } 求1+1/2+1/3+……+ 1/n #include <stdio.h> int main(void) { int i; double sum; printf(“Enter n: "); scanf("%d", &n); sum = 0; for ( i = 1; i <= n; i++ ) sum = sum + 1.0/i; printf( “sum = %f\n”, sum); return 0; }