The sum of the reciprocals 1+ 1/2 + 1/3 + 1/4 + . is infinite. Write a program that reads in a target and finds the first n such that 1+1/2+ 1/3+ . + 1n> target. reciprocal_sum.cpp 1 #include 3 using namespace std; int main() 6 { 7 double sum = ...; int n = ...; 10 11 12 13 cout <« "Target: " « endl; double target; cin >> target; while (...) { ... 17 18 19 } cout <« "n:" «n « endl; cout « "sum: " « sum « endl; 22 23 return 0;

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Please complete this in c++ . Modify code below. 

 

The sum of the reciprocals 1+ 1/2 + 1/3 + 1/4 + . is infinite. Write a program that reads in a target and finds the first n such that 1+ 1/2 + 1/3 + . + 1/n> target.
reciprocal_sum.cpp
1 #include <iostream>
using namespace std;
int main()
6 {
double sum = ...;
int n = ...;
cout « "Target: " « endl;
double target;
cin >> target;
while (...)
}
cout <« "n: " <«n « endl;
cout « "sum: " « sum <« endl;
22
return 0;
23 }
Transcribed Image Text:The sum of the reciprocals 1+ 1/2 + 1/3 + 1/4 + . is infinite. Write a program that reads in a target and finds the first n such that 1+ 1/2 + 1/3 + . + 1/n> target. reciprocal_sum.cpp 1 #include <iostream> using namespace std; int main() 6 { double sum = ...; int n = ...; cout « "Target: " « endl; double target; cin >> target; while (...) } cout <« "n: " <«n « endl; cout « "sum: " « sum <« endl; 22 return 0; 23 }
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