求函数f(n)=1/1!+1/2!+1/3!+...+1/n!的值

public Double Solution(int 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;
}