Calculating Pi with Taylor Method C++ and loop
<p>An approximate value of pi can be calculated using the series given below:</p>

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

<p>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.</p>

<p>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</p>

<p>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.</p>

<pre><code>#include &lt;stdio.h&gt;
#include &lt;iostream&gt;
#include &lt;cmath&gt;
using namespace std;

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

if (n &gt;0)
{

        double sum=1.0;
        int sign=-1;

    for (int i=1; i &lt;=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 &lt;&lt; "The approximation is " &lt;&lt; pi &lt;&lt; " using "&lt;&lt; n &lt;&lt; " terms" &lt;&lt; endl;
}
} while(n&gt;0);



    cout &lt;&lt; endl;

   return 0;
}
</code></pre>