$u$ harmonic and $\lim_{|z| \to \infty} u(z) = 0$ imply that $u \equiv 0$ in $\mathbb{C}$.

Question

If $u$ is harmonic and bounded in $\mathbb{C}$, then I've shown that $u$ is constant. I guess it can be helpful to show that $u \equiv 0$ if $u$ is harmonic and $\lim_{|z| \to \infty} u(z) = 0$, but how? I guess $u$ is not neccesary bounded in $\mathbb{C}$. What I can get is for all $\varepsilon > 0$ there exists $N \in \mathbb{N}$ such that $|u(z)| < \varepsilon$ if $|z| > N$. Any help is appreciated, thanks