给您一个级数1 + 1/2 ^ 2 + 1/3 ^ 3 +….. + 1 / n ^ n,找出直到第n个项的级数之和。
例子:
Input : n = 3
Output : 1.28704
Explanation : 1 + 1/2^2 + 1/3^3
Input : n = 5
Output : 1.29126
Explanation : 1 + 1/2^2 + 1/3^3 + 1/4^4 + 1/5^5
我们使用函数来计算功效。
C/C++
// C program to calculate the following series
#include
#include
// Function to calculate the following series
double Series(int n)
{
int i;
double sums = 0.0, ser;
for (i = 1; i <= n; ++i) {
ser = 1 / pow(i, i);
sums += ser;
}
return sums;
}
// Driver Code
int main()
{
int n = 3;
double res = Series(n);
printf("%.5f", res);
return 0;
}
Java
// Java program to calculate the following series
import java.io.*;
class Maths {
// Function to calculate the following series
static double Series(int n)
{
int i;
double sums = 0.0, ser;
for (i = 1; i <= n; ++i) {
ser = 1 / Math.pow(i, i);
sums += ser;
}
return sums;
}
// Driver Code
public static void main(String[] args)
{
int n = 3;
double res = Series(n);
res = Math.round(res * 100000.0) / 100000.0;
System.out.println(res);
}
}
Python
# Python program to calculate the following series
def Series(n):
sums = 0.0
for i in range(1, n + 1):
ser = 1 / (i**i)
sums += ser
return sums
# Driver Code
n = 3
res = round(Series(n), 5)
print(res)
C#
// C# program to calculate the following series
using System;
class Maths {
// Function to calculate the following series
static double Series(int n)
{
int i;
double sums = 0.0, ser;
for (i = 1; i <= n; ++i) {
ser = 1 / Math.Pow(i, i);
sums += ser;
}
return sums;
}
// Driver Code
public static void Main()
{
int n = 3;
double res = Series(n);
res = Math.Round(res * 100000.0) / 100000.0;
Console.Write(res);
}
}
/*This code is contributed by vt_m.*/
PHP
output:1.28704
输出:
1.28704