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

• Difficulty Level : Easy
• Last Updated : 24 May, 2021

Given a positive integer n, the problem is to find the sum of the given series of 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 ```

Naive Approach: Use a for loop to calculate each term iteratively and add it 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

 ``

## Javascript

 ``

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

 ``

## Javascript

 ``

Output :

`0.909091`

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