求函数f(n)=1/1!+1/2!+1/3!+…+1/n!的值

publi++c Double Solution(i++nt n)
{
Dounle ret = 0;
for(int i = 1;i <= n;i++)
{
int dishu = 1;
for(int j = 1;j <= n;j++)
{
dishu *= j;
}
ret += 1 / (Double)dishu;
}
return ret;
}