Perceptron - Convergence proof
What you presented is the  typical proof of convergence of perceptron proof indeed is independent of $\mu$. Hence the conclusion is right. 

Typically $\theta^*x$ represents a hyperplane that perfectly separate the two classes. 

The formula $k \le \frac{\mu^2 R^2 \|\theta^*\|^2}{\gamma^2}$ doesn't make sense as it implies that if you set $\mu$ to be small, then $k$ is arbitarily close to $0$. It is saying that with small learning rate, it converges immediately. You might want to look at the termination condition for your perceptron algorithm carefully.