数列求和

#include<stdio.h>
int Fuhao(i);
int main()
{
	int i,n;
	double total=0;
	printf("Please input n:");
	scanf("%d",&n);
	if(n<=0||n>=1000)
		printf("Input error!\n");
	else
	{
		for(i=1;i<=n;i++)
			total+=Fuhao(i)*(double)1/i;
	    printf("项数和为：%.2lf\n",total);
	}
	return 0;
}
int Fuhao(i)
{
	if(i%2==0)
		return -1;
}