Program to find the sum of series 1/1+1/2+1/3..+1/N

Problem statement:- Program to find the sum of series 1/1+1/2+1/3..+1/N

 Data requirement:-

   Input Data:- n

  Output Data:- sum

  Additional Data:- i

Program in C

  

Here is the source code of the C Program to find the sum of series 1/1+1/2+1/3..+1/N.

Code:

  

#include<stdio.h>
int main()
{
    int n;
    double sum=0.0,i;
    printf("Enter the range of number:");
    scanf("%d",&n);
    for(i=1;i<=n;i++)
        sum+=(1/i);
    printf("The sum of the series = %0.2f",sum);
}

Input/Output:

Enter the range of number:7
The sum of the series = 2.59

Program in C++

  

Here is the source code of the C++ Program to find the sum of series 1/1+1/2+1/3..+1/N.

Code:

  

#include<iostream>
#include<cmath>
using namespace std;
int main()
{
    int n;
    double sum=0.0,i;
    cout<<"Enter the range of number:";
    cin>>n;
    for(i=1;i<=n;i++)
         sum+=(1/i);
    cout<<"The sum of the series = "<<sum;
}

Input/Output:

Enter the range of number:9
The sum of the series = 2.82897

Program in Java

  

Here is the source code of the Java Program to find the sum of series 1/1+1/2+1/3..+1/N.

Code:

import java.util.Scanner;
public class p5 {
public static void main(String[] args) {
Scanner cs=new Scanner(System.in);
    int n;
    double sum=0.0,i;
    System.out.println("Enter the range of number:");
    n=cs.nextInt();
    for(i=1;i<=n;i++)
        sum+=(1/i);
    System.out.println("The sum of the series = "+sum);
    cs.close();
}
}

Input/Output:

Enter the range of number:
12
The sum of the series = 3.103210678210678

Program in Python

  

Here is the source code of the Python Program to find the sum of series 1/1+1/2+1/3..+1/N.

Code:

print("Enter the range of number:")
n=int(input())
print("Enter the value of x:");
x=int(input())
sum=0
i=1
while(i<=n):
    sum+=pow(x,i)
    i+=2
print("The sum of the series = ",sum)

Input/Output:

Enter the range of number:
5
The sum of the series =  2.283333333333333

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