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// 求1-1/2+1/3-1/4...+1/99-1/100
#include <stdio.h>
int main()
{
double a = 1.0;
int i;
double sum = 1.0;
for( i = 2; i <= 100; i++ )
{
a = (-1) * a;
sum = sum + a/i;
}
printf("1-1/2+1/3-1/4...+1/99-1/100=%f\n",sum);
return 0;
}
<img src="http://img.blog.csdn.net/20150423114517322?watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQvemhhb3lhcWlhbjU1Mg==/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/gravity/Center" alt="" />
【c语言】求1-1/2+1/3-1/4...+1/99-1/100
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原文地址：http://blog.csdn.net/zhaoyaqian552/article/details/45219715
