Implement two methods (using iterative and recursive approaches) to compute the sum of the reciprocals of the first n natural numbers. Mathematically, it can be denoted as follows: Sum(n) = 1 + 1 2+1 3 + ... + 1 n (1) Example 1: Input: [1] Output: [1.0] Explanation: n=1, so sum(1)=1.0 Example 2: Input: [2] Output: [1.5] Explanation: n = sum(2)=1+1/2=1.5 O Iterative Approach: public static float sum iterative (int n) { } 1 O Recursive Approach: Hint for recursive implementation: The prototype for the recursive approach should be as follows: Sum(n) = ( 1.0 if n == 1 Sum(n - 1) + 1/n if n > 1 (2) public static float sum recursive (int n) { = 2, so mas
Implement two methods (using iterative and recursive approaches) to compute the sum of the reciprocals of the first n natural numbers. Mathematically, it can be denoted as follows: Sum(n) = 1 + 1 2+1 3 + ... + 1 n (1) Example 1: Input: [1] Output: [1.0] Explanation: n=1, so sum(1)=1.0 Example 2: Input: [2] Output: [1.5] Explanation: n = sum(2)=1+1/2=1.5 O Iterative Approach: public static float sum iterative (int n) { } 1 O Recursive Approach: Hint for recursive implementation: The prototype for the recursive approach should be as follows: Sum(n) = ( 1.0 if n == 1 Sum(n - 1) + 1/n if n > 1 (2) public static float sum recursive (int n) { = 2, so mas
Computer Networking: A Top-Down Approach (7th Edition)
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Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
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![Implement two methods (using iterative and recursive
approaches) to compute the sum of the reciprocals of the first n
natural numbers. Mathematically, it can be denoted as follows:
Sum(n) = 1 + 1 2+1 3+ ... + 1 n (1) Example 1: Input: [1]
Output: [1.0] Explanation: n= 1, so sum(1)=1.0 Example 2:
Input: [2] Output: [1.5] Explanation: n
sum(2)=1+1/2=1.5 O Iterative Approach: public static float
sum iterative (int n) { } 1 0 Recursive Approach: Hint for
recursive implementation: The prototype for the recursive
approach should be as follows: Sum(n) = ( 1.0 if n== 1 Sum(n
- 1) + 1/n if n >1 (2) public static float sum recursive (int n) {
-
= 2, so](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F47bba70b-d35c-4545-add1-e8862c4403b1%2Faf48b64b-784c-459e-b904-6f794fafb0a3%2F8dmjb0c_processed.png&w=3840&q=75)
Transcribed Image Text:Implement two methods (using iterative and recursive
approaches) to compute the sum of the reciprocals of the first n
natural numbers. Mathematically, it can be denoted as follows:
Sum(n) = 1 + 1 2+1 3+ ... + 1 n (1) Example 1: Input: [1]
Output: [1.0] Explanation: n= 1, so sum(1)=1.0 Example 2:
Input: [2] Output: [1.5] Explanation: n
sum(2)=1+1/2=1.5 O Iterative Approach: public static float
sum iterative (int n) { } 1 0 Recursive Approach: Hint for
recursive implementation: The prototype for the recursive
approach should be as follows: Sum(n) = ( 1.0 if n== 1 Sum(n
- 1) + 1/n if n >1 (2) public static float sum recursive (int n) {
-
= 2, so
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