1/2+1/6+1/12+1/20+.......+1/n(n+1)

这是中学的奥赛题啦，不是用电脑算的啦。
上题解答如下
=(2-1)/1*2+(3-2)/2*3+(4-3)/3*4+(5-4)/4*5...+(n+1-n)/n*(n+1)
=2/1*2-1/1*2+3/2*3-2/2*3+4/3*4-3/3*4+5/4*5-4/4*5..+(n+1)/n(n+1)-n/n(n+1)
=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5...+1/n-1/(n+1)
=1-1/(n+1)

要做什么？

double fun(int n)
{
if(n<=0)
return 0;
double sum =0.00000;
for(int i=1;i<=n;i++)
{
sum += 1/(i*(i+1))
}
}

C语言是这样做的:
#include "stdio.h"
void main()
{
int i,n;
double sum=0.0;
printf("Input a number:");
scanf("%d",&n);
if(n<=0)
printf("The number error!");
else
for(i=1;i<=n;i++) sum+=1.0/(i*(i+1));/*楼上这个地方有点小错还有忘了返回值纠正下*/
printf("1/2+1/6+1/12+1/20+.......+1/%d(%d+1)=%f",n,n,sum);

}

#include "stdio.h"

main()
{
double a;

printf("input n\n");

scanf("lf",&a);
<