Program to find sum of series 1 + 1/2 + 1/3 + 1/4 + .. + 1/n
If inverse of a sequence follows rule of an A.P i.e, Arithmetic progression, then it is said to be in Harmonic Progression.In general, the terms in a harmonic progression can be denoted as : 1/a, 1/(a + d), 1/(a + 2d), 1/(a + 3d) …. 1/(a + nd).
As Nth term of AP is given as ( a + (n – 1)d) .Hence, Nth term of harmonic progression is reciprocal of Nth term of AP, which is : 1/(a + (n – 1)d)
where “a” is the 1st term of AP and “d” is the common difference.
We can use a for loop to find sum.
C++
#include <iostream>
using namespace std;
class gfg
{
public : double sum(int n)
{
double i, s = 0.0;
for (i = 1; i <= n; i++)
s = s + 1/i;
return s;
}
};
int main()
{
gfg g;
int n = 5;
cout << "Sum is " << g.sum(n);
return 0;
}
|
C
#include <stdio.h>
double sum(int n)
{
double i, s = 0.0;
for (i = 1; i <= n; i++)
s = s + 1/i;
return s;
}
int main()
{
int n = 5;
printf("Sum is %f", sum(n));
return 0;
}
|
Java
import java.io.*;
class GFG {
static double sum(int n)
{
double i, s = 0.0;
for (i = 1; i <= n; i++)
s = s + 1/i;
return s;
}
public static void main(String args[])
{
int n = 5;
System.out.printf("Sum is %f", sum(n));
}
}
|
Python3
def sum(n):
i = 1
s = 0.0
for i in range(1, n+1):
s = s + 1/i;
return s;
n = 5
print("Sum is", round(sum(n), 6))
|
C#
using System;
class GFG {
static float sum(int n)
{
double i, s = 0.0;
for (i = 1; i <= n; i++)
s = s + 1/i;
return (float)s;
}
public static void Main()
{
int n = 5;
Console.WriteLine("Sum is "
+ sum(n));
}
}
|
PHP
<?php
function sum( $n)
{
$i;
$s = 0.0;
for ($i = 1; $i <= $n; $i++)
$s = $s + 1 / $i;
return $s;
}
$n = 5;
echo("Sum is ");
echo(sum($n));
?>
|
Output:
2.283333
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