I am having trouble proving the following statement:
Let $\{X_n\}$ be a sequence of independent random variables such that $E[X_n] = 0$, $E[X_n^2] = 1$ and $E[|X_n|^{2+\delta}] < M$ for some $\delta > 0, M \in \mathbb{R}^+$. Prove that: $$\frac{1}{\sqrt{\ln n}}\sum_{k=1}^n \frac{X_k}{\sqrt{k}}\Rightarrow N(0,1)$$ I think I need to use the Lindberg Feller CLT to prove it, but I'm not sure. Could someone please provide some guidance?
Apply the Lyapunov CLT to $Y_n:=\frac{X_n}{\sqrt n}$. The $Y_n$ are independent, $E(Y_n)=0$ and $V(Y_n) = \frac 1n$, so that $D_n:=\sqrt{\sum_{k=1}^n V(Y_k)} = \sqrt{\sum_{k=1}^n \frac 1k}\sim \sqrt{\log n}$.
Lyapunov's condition is verified: $$\frac{1}{D_n^{2+\delta}} \sum_{k=1}^n E\left[| Y_k|^{2+\delta}\right]=O\left( \frac{1}{(\log n)^{1+\delta/2}} \sum_{k=1}^n \frac{1}{k^{1+\delta/2}}\right) = O\left(\frac{1}{(\log n)^{1+\delta/2}}\right)\xrightarrow[n\to \infty]{} 0 $$
Hence $$\frac{S_n-E(S_n)}{\sqrt{V(S_n)}} =\frac 1{\sqrt{\sum_{k=1}^n \frac 1k}} \sum_{k=1}^n \frac{X_k}{\sqrt k} \xrightarrow{d} N(0,1)$$
Finally, write $$\frac 1{\sqrt{\log n}} \sum_{k=1}^n \frac{X_k}{\sqrt k} = \frac{\sqrt{\sum_{k=1}^n \frac 1k}}{\sqrt{\log n}} \left(\frac 1{\sqrt{\sum_{k=1}^n \frac 1k}} \sum_{k=1}^n \frac{X_k}{\sqrt k}\right)$$ and use Slutsky's theorem to conclude that $$\frac 1{\sqrt{\log n}} \sum_{k=1}^n \frac{X_k}{\sqrt k}\xrightarrow{d} N(0,1)$$