# Sum of the Series 1/(1*2) + 1/(2*3) + 1/(3*4) + 1/(4*5) + . . . . .

Given a positive integer n, The problem is to find the sum of the given series upto n terms:
1/(1*2) + 1/(2*3) + 1/(3*4) + 1/(4*5) + . . . . . . . + 1/(n*(n+1))

Examples :

```Input : 3
Output : 0.75
( 1/(1*2)+ 1/(2*3) + 1/(3*4) )
= (1/2 + 1/6 + 1/12)
= 0.75

Input : 10
Output : 0.909
( 1/(1*2) + 1/(2*3) + 1/(3*4) + 1/(4*5) +
1/(5*6) + 1/(6*7) + 1/(7*8) + 1/(8*9) +
1/(9*10) + 1/(10*11) )
= (1/2 + 1/6 + 1/12 + 1/20 + 1/30 +
1/42 + 1/56 + 1/72 + 1/90 + 1/110)
= 0.909
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach: Use a for loop to calculate each term iteratively and add to the final sum.

## C++

 `// C++ program to find the sum of given series ` `#include ` `using` `namespace` `std; ` ` `  `// function to find the sum of given series ` `double` `sumOfTheSeries(``int` `n) ` `{ ` `    ``// Computing sum term by term ` `    ``double` `sum = 0.0; ` `    ``for` `(``int` `i = 1; i <= n; i++) ` `        ``sum += 1.0 / (i * (i + 1));    ` `    ``return` `sum; ` `} ` ` `  `// driver program to test above function ` `int` `main() ` `{ ` `    ``int` `n = 10; ` `    ``cout << sumOfTheSeries(n); ` `    ``return` `0; ` `} `

## Java

 `// Java program to find the sum of given series ` `class` `demo { ` ` `  `    ``// function to find the sum of given series ` `    ``public` `static` `double` `sumOfTheSeries(``int` `n) ` `    ``{ ` `       ``// Computing sum term by term ` `        ``double` `sum = ``0.0``; ` `        ``for` `(``int` `i = ``1``; i <= n; i++)  ` `            ``sum += ``1.0` `/ (i * (i + ``1``)); ` `        ``return` `sum; ` `    ``} ` ` `  `    ``// driver program to test above function ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``int` `n = ``10``; ` `        ``System.out.println(sumOfTheSeries(n)); ` `    ``} ` `} `

## Python3

 `# Python3 code to find the sum of given series ` ` `  `# Function to find the sum of given series ` `def` `sumOfTheSeries( n ): ` `     `  `    ``# Computing sum term by term ` `    ``sum` `=` `0` `    ``for` `i ``in` `range``(``1``, n ``+` `1``): ` `        ``sum` `+``=` `1.0` `/` `(i ``*` `(i ``+` `1``));  ` `    ``return` `sum` ` `  ` `  `# Driver function ` `if` `__name__ ``=``=` `'__main__'``: ` `     `  `    ``ans ``=` `sumOfTheSeries(``10``) ` `     `  `    ``# Rounding decimal value to 6th decimal place ` `    ``print` `(``round``(ans, ``6``)) ` ` `  `# This code is contributed by 'saloni1297' `

## C#

 `// C# program to find the sum of given series ` `using` `System; ` ` `  `class` `demo { ` ` `  `    ``// Function to find the sum of given series ` `    ``public` `static` `double` `sumOfTheSeries(``int` `n) ` `    ``{ ` `        ``// Computing sum term by term ` `        ``double` `sum = 0.0; ` `        ``for` `(``int` `i = 1; i <= n; i++)  ` `            ``sum += 1.0 / (i * (i + 1)); ` `        ``return` `sum; ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 10; ` `        ``Console.Write(sumOfTheSeries(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m `

## PHP

 ` `

Output :

```0.909091
```

Efficient Approach: Use the formula n/(n+1)

```Validity of the formula:
Sum upto n terms = 1/(1*2) + 1/(2*3) + 1/(3*4) +
........ + 1/(n*(n+1))
where
1st term = 1/(1*2)
2nd term = 1/(2*3)
3rd term = 1/(3*4)
.
.
.
.
n-th term = 1/(n*(n+1))

i.e. the k-th term is of the form 1/(k*(k+1))
which can further be written as k-th term =
1/k - 1/(k+1)

So sum upto n terms can be calculated as:
(1/1 - 1/1+1) + (1/2 - 1/2+1) + (1/3 - 1/3+1)
+ ......... + (1/n-1 - /1n) + (1/n - 1/n+1)
= (1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + .........
+ (1/n-1 - 1/n) + (1/n - 1/n+1)
= 1 - 1/n+1
= ((n+1) - 1)/n+1
= n/n+1

Hence sum  upto n terms = n/n+1
```

## C++

 `// C++ program to find sum of given series ` `#include ` `using` `namespace` `std; ` ` `  `// function to find sum of given series ` `double` `sumOfTheSeries(``int` `n) ` `{ ` `    ``// type-casting n/n+1 from int to double ` `    ``return` `(``double``)n / (n + 1); ` `} ` ` `  `// driver program to test above function ` `int` `main() ` `{ ` `    ``int` `n = 10; ` `    ``cout << sumOfTheSeries(n); ` `    ``return` `0; ` `} `

## Java

 `// Java program to find sum of given series ` `class` `demo { ` ` `  `    ``// function to find sum of given series ` `    ``public` `static` `double` `sumOfTheSeries(``int` `n) ` `    ``{ ` `        ``// type -casting n/n+1 from int to double ` `        ``return``(``double``)n / (n + ``1``); ` `    ``} ` ` `  `    ``// driver program to test above function ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``int` `n = ``10``; ` `        ``System.out.println(sumOfTheSeries(n)); ` `    ``} ` `} `

## Python3

 `# Python3 code to find sum of given series ` ` `  `# Function to find sum of given series ` `def` `sumOfTheSeries(n): ` `     `  `    ``# Type-casting n/n+1 from int to float ` `    ``return` `(``float``(n) ``/` `(n ``+` `1``)) ` ` `  `# Driver function    ` `if` `__name__ ``=``=` `'__main__'``: ` `         `  `    ``n ``=` `10` `    ``ans ``=` `sumOfTheSeries(n) ` `     `  `    ``# Rounding decimal value ` `    ``print` `(``round``(ans, ``6``)) ` ` `  `# This code is contributed by 'saloni1297' `

## C#

 `// C# program to find sum of given series ` `using` `System; ` ` `  `class` `demo { ` ` `  `    ``// Function to find sum of given series ` `    ``public` `static` `double` `sumOfTheSeries(``int` `n) ` `    ``{ ` `        ``// type -casting n/n+1 from int to double ` `        ``return``(``double``)n / (n + 1); ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 10; ` `        ``Console.Write(sumOfTheSeries(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 ` `

Output :

```0.909091
```

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