The sum of the reciprocals 1 + 1/2 + 1/3 + 1/4 + … is infinite. I’m supposed to write a program that reads in a target and finds the first n such that 1 + 1/2 + 1/3 + … + 1/n > target, but I can’t figure out how to do it. This is where I’m currently at, and the following code just results in an infinite loop. Any help will be greatly appreciated. Doublelnvestment2.java XReciprocalSum.java XCompare2.java XTaxCalculator.java XaPro 1import java.util.Scanner;3./**4This program computes how many steps the sum 1 + 1/2 + 1/3 + ...needs to exceed a given target.vestme6.SimulatA */7 ►public class ReciprocalSum8.{6.public static void main (String[] args)rinter10{Demo11procalSum 12pingCalculatorume1double sum = 0;13int n = 0;1415Scanner in= new Scanner (System.in);350.imlSystem.out.print("Target: ");double target = in.nextDouble();16Libraries17es and Conso1819while (sum < target) {28n++;21sum = sum + (1 / n);|222324System.out.println("n: " + n);25System.out.println("sum: " + sum);26278 TODOMacBook Pro

The sum of the reciprocals 1 + 1/2 + 1/3 + 1/4 + … is infinite. I’m supposed to write a program that reads in a target and finds the first n such that 1 + 1/2 + 1/3 + … + 1/n > target, but I can’t figure out how to do it. This is where I’m currently at, and the following code just results in an infinite loop. Any help will be greatly appreciated.

The sum of the reciprocals 1 + 1/2 + 1/3 + 1/4 + … is infinite. I’m supposed to write a program that reads in a target and finds the first n such that 1 + 1/2 + 1/3 + … + 1/n > target, but I can’t figure out how to do it. This is where I’m currently at, and the following code just results in an infinite loop. Any help will be greatly appreciated.

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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