Programa para hallar la suma de una Serie 1 + 1/2^2 + 1/3^3 + …..+ 1/n^n

Te han dado una serie 1 + 1/2^2 + 1/3^3 + …..+ 1/n^n, encuentra la suma de la serie hasta el término n.

Ejemplos:

Entrada: n = 3

Salida: 1.28704

Explicación: 1 + 1/2^2 + 1/3^3

Entrada: n = 5

Salida: 1.29126

Explicación: 1 + 1/2^2 + 1/3^3 + 1/4^4 + 1/5^5

Usamos la función de potencia para calcular la potencia.

// C++ program to calculate the following series
#include <bits/stdc++.h>
using namespace std;
 
// 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);
     
    cout << res;
    return 0;
}
 
// This code is contributed by Ankita saini

// C program to calculate the following series
#include <math.h>
#include <stdio.h>
   
// 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 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 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# 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
// PHP program to calculate
// the following series
   
// Function to calculate
// the following series
function Series($n)
{
    $i;
    $sums = 0.0;
    $ser;
    for ($i = 1; $i <= $n; ++$i)
    {
        $ser = 1 / pow($i, $i);
        $sums += $ser;
    }
    return $sums;
}
   
    // Driver Code
    $n = 3;
    $res = Series($n);
    echo $res;
   
// This code is contributed by Vishal Tripathi.
?>

<script>
 
// Javascript program to calculate
// the following series
function Series(n)
{
    let sums = 0.0;
     
    for(let i = 1; i < n + 1; i++)
    {
        ser = 1 / Math.pow(i, i);
        sums += ser;
    }
    return sums;
}
 
// Driver Code
let n = 3;
let res = Math.round(Series(n) * 100000) / 100000;
 
document.write(res);
 
// This code is contributed by rohitsingh07052
 
</script>

Producción: 

1.28704

Complejidad de tiempo: O(n)

Espacio Auxiliar: O(1)