Does $\int_{|z|<\frac{1}{n}} \frac{1}{|z|^{2(d-\alpha)}} \to 0$

Question

I am wondering if $\int_{|z|<\frac{1}{n}} \frac{1}{|z|^{2(d-\alpha)}} \to 0$ as $n \to \infty$ for $z \in \mathbb{R}^d$. It is known that $\int_{\mathbb{R}^d} \frac{1}{|z|^{d-\alpha}}$ is Lebesgue integrable whenever $\alpha > 0$, i.e. when $d-\alpha < d$. I'm not even certain whether $\int_{|z|<\frac{1}{n}} \frac{1}{|z|^{2(d-\alpha)}}$ is square integrable, let alone converges to $0$. What happens to this integral as we integrate over the disk of radius $\frac{1}{n}$?