Calculating Pi with Taylor Method C++ and loop

Question

An approximate value of pi can be calculated using the series given below:

pi = 4 * [ 1 - 1/3 + 1/5 - 1/7 + 1/9 … + ((-1)^n)/(2n + 1) ]

write a C++ program to calculate the approximate value of pi using this series. The program takes an input n that determines the number of terms in the approximation of the value of pi and outputs the approximation. Include a loop that allows the user to repeat this calculation for new values n until the user says she or he wants to end the program.

The expect result is: Enter the number of terms to approximate (or zero to quit): 1 The approximation is 4.00 using 1 term. Enter the number of terms to approximate (or zero to quit): 2 The approximation is 2.67 using 2 terms. Enter the number of terms to approximate (or zero to quit): 0

I can get the correct result now but I don't know how to Include a loop that allows the user to repeat this calculation for new values n until the user says she or he wants to end the program.

#include <stdio.h>
#include <iostream>
#include <cmath>
using namespace std;

int main()
{ int n; double pi, sum; 
do { 
     cout << "Enter the number of terms to approximate (or zero to quit):" << endl;
     cin >> n;

if (n >0)
{

        double sum=1.0;
        int sign=-1;

    for (int i=1; i <=n; i++) {
    sum += sign/(2.0*i+1.0);
        sign=-sign;
    }
        pi=4.0*sum;

    cout.setf(ios::fixed);
    cout.setf(ios::showpoint);
    cout.precision(2);

    cout << "The approximation is " << pi << " using "<< n << " terms" << endl;
}
} while(n>0);



    cout << endl;

   return 0;
}

Answer 1

You have wrong initialization:

double sum=0.0;
int sign=1;

It should be

double sum = 1.0;
int sign = -1;

The loop is also wrong (has a typo?), it should be

for (int i = 1; i < n; i++) { /* please, notice "i < n" and "{"  */
    sum += sign / (2.0 * i + 1.0);
    sign = -sign; /* more readable, IMHO than  sign *=-1; */
}

pi = 4.0 * sum; /* you can well move it out of the loop */

EDIT if you want to repeat calculation a common practice is extrating a function (do not cram everything into single main):

double compute(int n) {
  if (n < 0) 
    return 0.0;

  double sum = 1.0;
  int sign = -1;

  for (int i = 1; i < n; i++) { 
    sum += sign / (2.0 * i + 1.0);
    sign = -sign; /* more readable, IMHO than  sign *=-1; */
  }

  return 4.0 * sum;
}

EDIT 2 the main function could be something like this:

int main() {
  int n = -1;

  /* quit on zero */
  while (n != 0) {
    cout << "Enter the number of terms to approximate (or zero to quit):" << endl;
    cin >> n;

    if (n > 0)
      cout << "The approximation is " << compute(n) << " using "<< n << " terms" << endl;
  }
}

Answer 2

The alternating sign must be part of your loop. Include it into the loop body by using a compound statement:

for (int i=1; i <=n; i++) {
  sum += sign/(2.0*i+1.0);
  sign *=-1;
}