Say I have a sequence of i.i.d random variables $(X_n)$ with mean $\mu< \infty $ and variance $\sigma^2 < \infty$.
Using the CLT, I can state that:
$\lim_{n\rightarrow\infty} \mathbb{P}\bigg( \frac{\sum_{i=1}^n{X_i} - n\mu}{\sigma n^{1/2}} \leqslant x\bigg) = \Phi(x)$
where $\Phi$ is the CDF of a standard normal distribution.
I'm wondering if there is a simple manipulation of this identify to obtain the value of the following limit in terms of the function $\Phi$:
$\lim_{n\rightarrow\infty} \mathbb{P}\bigg( \frac{\sum_{i=1}^n{X_i} - n\mu}{\sigma n^{\alpha}} \leqslant x\bigg).$
Note that the only difference between the CLT and the second equation is that we are now dividing by $n^\alpha$ instead of $n^{1/2}$. In addition, we can assume that $\alpha >0$.