Having written the following program in Python:
#Collatz Conjecture
#if n is odd, 3n+1
#if n is even, n/2
while True:
print("**********")
n=input("Which number would you like to test? ")
i=1
print("**********")
while n!=1:
print("Step " + str(i) + ": " + str(n))
n=float(n)
if(n%2==1):
n=(n*3)+1
elif(n%2==0):
n=n/2
else:
print("Invalid number.")
i+=1
When given the number: 111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111
The second step is the result inf - does this break the Collatz Conjecture, or not?
This is an artifact of programming; in particular, you got that result because the number provided is larger than $2^{128}$, hence a single-precision floating point number cannot represent it and rounds it to infinity. Obviously, it it is not possible for the Collatz iteration to ever reach the value infinity.
Also, as a note, I don't know why you have the line
n = (float)n, but such a cast is a really bad idea, since you are dealing with exact computation in integers, and this line will cause rounding issues for larger integers. Rather, you need to use an arbitrarily sized integer type for this sort of computation.