My switch statement embedded in my do loop doesn't seem to execute correctly:

Instead of correctly recognizing the "#" term approximation, it prints a blank line. Anyone have some input?Code:#include <stdio.h> #include <math.h> int main() { double x, y, y2; int n_term; printf("Enter -1<x<1 : "); scanf("%lf",&x); y = log (x); printf("\nTrue value of log(1+x) = %.5f\n\n",y); do { printf("Enter an integer 1-4; 0 to Exit\n"); scanf("%d",&n_term); switch (n_term){ case '1': printf("\n1 term approximation\n"); y2 = x; printf("Approximate log(1+x) = %.5f",y2); break; case '2': printf("\n2 term approximation\n"); y2 = x - (pow(x,2)/2); printf("Approximate log(1+x) = %.5f",y2); break; case '3': printf("\n2 term approximation\n"); y2 = x - (pow(x,2)/2) + (pow(x,3)/3); break; printf("Approximate log(1+x) = %.5f",y2); case '4': y2 = x - (pow(x,2)/2) + (pow(x,3)/3) - (pow(x,4)/4); break; case '0': break; default: printf("unrecognized operator"); break; } } while (n_term != 0); scanf("%lf",&x); }