Program to find the sum of a Series 1 + 1/2^2 + 1/3^3 + …..+ 1/n^n

You have been given a series 1 + 1/2^2 + 1/3^3 + …..+ 1/n^n, find out the sum of the series till nth term.
Examples:

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

We use use power function to compute power.

C/C++

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// 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;
}

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Java














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// 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); 


    } 


} 



















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Python














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# 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) 



















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C#














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// 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.*/ 



















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PHP














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<?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. 


?> 



















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output:


1.28704





Output:


1.28704





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Improved By :  vt_m


					
		

    Article Tags : 
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Practice Tags : 
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