C++ Infinite Sum Code Hanging? Freezing? I'm not sure

Question

Okay, so I'm trying to compute an infinite sum using a for loop instead of a while loop, and the darn code isn't working and I'm not sure why. I get the two "I got here"'s ONCE and then nothing. :\

Does anyone know why my code isn't working and just looks like it freezes or hangs? The loops should end with the way I made them, right? Here's the current problem code

        check=0;
        cerr << "I got here" << endl;
        for (int m=1;m<m+1;m++)
        {
          cerr << "I got here" << endl;
            for (int n=1;n<n+1;n++)
            {
                if (check==0)
                {
                  anaPhi[i][j][k]=1.0/(m*m*m*n*n*n)
                  *cos(k*deltat*sqrt(5.0)*pi/4.0
                  *sqrt(m*m+4.0*n*n))
                  *sin(m*pi*i*h/4.0)
                  *sin(n*pi*j*h/2.0);
                  check=1;
                }
                else
                {
                  anaPhi[i][j][k]= anaPhi[i][j][k]
                  +1.0/(m*m*m*n*n*n)
                  *cos(k*deltat*sqrt(5.0)*pi/4.0
                  *sqrt(m*m+4.0*n*n))
                  *sin(m*pi*i*h/4.0)
                  *sin(n*pi*j*h/2.0);
                }
            }
        }
    }

I just tested this code and this code also doesn't work!

#include <iostream>
#include <cmath>
using namespace std;

int main()
{
    int b=0;
    for (int m=1;m<m+1;m=m+2)
    {
        b=cos(b)+sqrt(b);
    }
    cout << b << endl;
    return 0;
}

Answer 1

I see a couple of issues.

1. if(check=0) will always set check to 0.

2. Because you use n= n+2, it will not ever break. Because at some point you get to n = 0x7FFFFFFE (highest positive even number) and in this case n+1 is > n. It then wraps to 0x80000000 (the most negative signed int, still even) and n+1 will still be > n. If you were able to get n = 0x7FFFFFFF, then n + 1 would actually be less than n due to integer wrap around.

3. There does not seem to be any other way to break out of the for loops than meeting the coditions set.