## C++ program to find the sum of 1 + 1/2 + 1/3 + 1/4+…n:

In this *C++* programming tutorial, we will learn how to find the sum of the series *1 + 1/2 + 1/3 + 1/4 + …* upto *nth* term. This program will take the value of *n* as an *input* from the user and it will print the *sum*.

With this program, you will learn how to *read user inputs*, *how to use loop* and how to do simple *mathematical calculations* in *C++*.

### Algorithm to solve this problem:

If you look closely to the series *1 + 1/2 + 1/3 + 1/4…*, the series is actually is the *sum* of the inverse of *1, 2, 3, 4,…* etc. So, we can use a *loop* to run from *1* to *n* and in each iteration, we can add its inverse to a *sum* variable.

This series is also called a *Harmonic progression*. *Harmonic progression* is a series, the element of which is the *reverse* of an *arithmetic progression*. For example, if *n1, n2, n3, n4, n5…* is an *arithmetic progression*, *1/n1, 1/n2, 1/n3…* is a *harmonic progression*.

In our case, it is a *Harmonic progression* series for the *arithmetic progression* *1, 2, 3, 4, 5…*.

- Initialize one
*sum*variable as*0*to hold the sum. - Use a
*loop*to iterate from*i = 1*to*i = n*. - In each iteration, add the value of
*1/i*to*sum*. - Finally, print the
*sum*.

## C++ program:

Below is the complete *C++ program* that finds the *sum* of the series *1 + 1/2 + 1/3 +…* upto *nth* term:

```
#include <iostream>
using namespace std;
float findSum(int n)
{
float sum = 0;
for (float i = 1; i <= n; i++)
{
sum += 1 / i;
}
return sum;
}
int main()
{
int n;
cout
<< "Enter the value of n: "
<< endl;
cin >> n;
cout << "Sum: " << findSum(n) << endl;
}
```

Here,

*n*is an*integer variable*to hold the value of*n*that is entered by the user.- By using
*cin*, we are reading the value of*n*. -
The value of

*sum*is calculated by using the*findSum*method. This method takes the value of*n*as its parameter and returns the sum, which is a*floating point*value.- In
*findSum*, we have initialized one*sum*variable as*0*, which is a*float value*. - The
*for loop*runs from*i = 1*to*i = n*. Inside this loop, it is adding the value of*1/i*to*sum*in each iteration. - Once the loop ends, it returns
*sum*, which holds the required sum.

- In

### Sample output:

If you run the above program, it will print output as like below:

```
Enter the value of n:
4
Sum: 2.08333
Enter the value of n:
6
Sum: 2.45
Enter the value of n:
11
Sum: 3.01988
```