Related Articles

Related Articles

Program to find the sum of a Series 1 + 1/2^2 + 1/3^3 + …..+ 1/n^n
  • Difficulty Level : Basic
  • Last Updated : 11 May, 2018

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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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);
    }
}

chevron_right


Python

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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)

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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.*/

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?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.
?>

chevron_right


output:

1.28704


Output:

1.28704

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up
Recommended Articles
Page :