# Program to find the sum of a Series 1 + 1/2^2 + 1/3^3 + …..+ 1/n^n

• Difficulty Level : Easy
• Last Updated : 04 Aug, 2022

You have been given a series 1 + 1/2^2 + 1/3^3 + …..+ 1/n^n, find out the sum of the series till nth term.

Examples:

```Input : n = 3
Output : 1.28704
Explanation : 1 + 1/2^2 + 1/3^3```
```Input : n = 5
Output : 1.29126
Explanation : 1 + 1/2^2 + 1/3^3 + 1/4^4 + 1/5^5```

We use power function to compute power.

## C++

 `// C++ program to calculate the following series``#include ``using` `namespace` `std;` `// Function to calculate the following series``double` `Series(``int` `n)``{``    ``int` `i;``    ``double` `sums = 0.0, ser;``    ``for``(i = 1; i <= n; ++i)``    ``{``        ``ser = 1 / ``pow``(i, i);``        ``sums += ser;``    ``}``    ``return` `sums;``}``  ` `// Driver Code``int` `main()``{``    ``int` `n = 3;``    ``double` `res = Series(n);``    ` `    ``cout << res;``    ``return` `0;``}` `// This code is contributed by Ankita saini`

## C

 `// C program to calculate the following series``#include ``#include ``  ` `// Function to calculate the following series``double` `Series(``int` `n)``{``    ``int` `i;``    ``double` `sums = 0.0, ser;``    ``for` `(i = 1; i <= n; ++i) {``        ``ser = 1 / ``pow``(i, i);``        ``sums += ser;``    ``}``    ``return` `sums;``}``  ` `// Driver Code``int` `main()``{``    ``int` `n = 3;``    ``double` `res = Series(n);``    ``printf``(``"%.5f"``, res);``    ``return` `0;``}`

## Java

 `// Java program to calculate the following series``import` `java.io.*;``  ` `class` `Maths {``  ` `    ``// Function to calculate the following series``    ``static` `double` `Series(``int` `n)``    ``{``        ``int` `i;``        ``double` `sums = ``0.0``, ser;``        ``for` `(i = ``1``; i <= n; ++i) {``            ``ser = ``1` `/ Math.pow(i, i);``            ``sums += ser;``        ``}``        ``return` `sums;``    ``}``  ` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `n = ``3``;``        ``double` `res = Series(n);``        ``res = Math.round(res * ``100000.0``) / ``100000.0``;``        ``System.out.println(res);``    ``}``}`

## Python3

 `# Python program to calculate the following series``def` `Series(n):``    ``sums ``=` `0.0``    ``for` `i ``in` `range``(``1``, n ``+` `1``):``        ``ser ``=` `1` `/` `(i``*``*``i)``        ``sums ``+``=` `ser``    ``return` `sums``  ` `# Driver Code``n ``=` `3``res ``=` `round``(Series(n), ``5``)``print``(res)`

## C#

 `// C# program to calculate the following series``using` `System;``  ` `class` `Maths {``  ` `    ``// Function to calculate the following series``    ``static` `double` `Series(``int` `n)``    ``{``        ``int` `i;``        ``double` `sums = 0.0, ser;``        ``for` `(i = 1; i <= n; ++i) {``            ``ser = 1 / Math.Pow(i, i);``            ``sums += ser;``        ``}``        ``return` `sums;``    ``}``  ` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``        ``int` `n = 3;``        ``double` `res = Series(n);``        ``res = Math.Round(res * 100000.0) / 100000.0;``        ``Console.Write(res);``    ``}``}``/*This code is contributed by vt_m.*/`

## PHP

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## Javascript

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Output:

`1.28704`

Time Complexity: O(nlogn) since using inbuilt pow function in loop

Auxiliary Space: O(1)

My Personal Notes arrow_drop_up