如何通过无穷级数循环常数e

Question

𝒆的值由无穷级数给出𝒆=𝟏+𝟏/𝟏！+𝟏/𝟐！+𝟏/𝟑！+𝟏/𝟒！+⋯上述算法在计算上是有效的。用它来计算𝒆到任意精度。

如何在公式中输入公式？

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[])
{
    double e;
    double n;
    float i;
    n = atof(argv[1]);
    e = 1;
    for (I = 0; I < n; i++)
    e = 1 + 1/n!;
    printf("%.10f\n", e);
    return 0;
}

Answer 1

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[])
{
    double e = 1.0, f = 1.0; // f is your 'accumulated' factorial
    int i, n = atoi(argv[1]); // Best to use integers for loop counting
    for (i = 1; i < n; i++) {
        f *= i; // First time, this wil be 1 * 1 = 1; second time 2, third 6, etc
        e += 1.0/f; // Divide here by our 'running' value for i!
    }
    printf("%.10f\n", e);
    return 0;
}

Answer 2

 long long factorial(int n){
    long long fact=1;
    for(i=2;i<=;i++)
     fact*=1;
     return fact;
    }

对于累积总和，您应该使用e = e + 1 + 1 / n！你可以计算出n！使用上述阶乘函数。如果您有任何疑问，请告诉我