Computing 1 + 1/3 + 1/5 +... + 1/(2n + 1) Value

The while loop is required and must be calculated to 1/(2n + 1) <0.00001.

Public class dowhiledemo
{
Public static void main (string ARGs [])
{
Int n = 1;
Double dsum= 1.0, dtemp;
Do
{
N = 2 * n + 1;
Dtemp = 1.0/N; // is critical. If 1.0 is written as an integer 1, the calculation result is incorrect. The division here is not calculated as a double-precision type, but as an integer division operation.
Dsum + = dtemp;
}
While (dtemp> = 0.00001 );
System. Out. println ("the value of N at the end of the loop is:" + n );
System. Out. println ("the calculated result is:" + dsum );
}
}