Application of Lyapunov's Central Limit Theorem (1)

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Let $S_n=\sum_{k=1}^nX_{nk}$ and $\{X_{nk}\}$ be independent random variables with $$\mathscr{P}\{X_{nk}=1\}=\frac{1}{n-k+1}=1-\mathscr{P}\{X_{nk}=0\}.$$We have $$\begin{array}{rl}
\alpha_{nk} & = \mathscr{E}(X_{nk}) = \frac{1}{n-k+1}; \\
\sigma^2_{nk} & = \mathscr{E}(X_{nk}-\alpha_{nk})^2 = \frac{1}{n-k+1}\left(1-\frac{1}{n-k+1}\right); \\
\gamma_{nk} & = \mathscr{E}|X_{nk}-\alpha_{nk}|^3 \\
& = \mathscr{E}(X_{nk}^3)-3\alpha_{nk}\mathscr{E}(X_{nk}^2)+3\alpha_{nk}^2\mathscr{E}(X_{nk})-\alpha_{nk}^3\\
&=\frac{1}{n-k+1}-\frac{3}{(n-k+1)^2}+\frac{3}{(n-k+1)^3}-\frac{1}{(n-k+1)^3}\\
&=\frac{(n-k+1)^2-3(n-k+1)+2}{(n-k+1)^3}.
\end{array}$$Let $$\begin{array}{rl}
\sigma^2_n &=\sigma^2(S_n)=\sum_{k=1}^n\sigma^2_{nk}=\sum_{k=1}^n\frac{n-k}{(n-k+1)^2}=\sum_{k=1}^n\frac{k-1}{k^2}; \\
\Gamma_n &=\sum_{k=1}^n\gamma_{nk}=\sum_{k=1}^n\frac{(n-k+1)^2-3(n-k+1)+2}{(n-k+1)^3}=\sum_{k=1}^n\frac{k^2-3k+2}{k^3}.
\end{array}$$By Lyapunov's condition, $$
\frac{1}{\sigma_n^3}\Gamma_n=\frac{\sum_{k=1}^n\frac{k^2-3k+2}{k^3}}{\left(\sum_{k=1}^n\frac{k-1}{k^2}\right)^{3/2}}=\frac{\sum_{k=1}^n\frac{1}{k}-3\sum_{k=1}^n\frac{1}{k^2}+2\sum_{k=1}^n\frac{1}{k^3}}{\left(\sum_{k=1}^n\frac{1}{k}-\sum_{k=1}^n\frac{1}{k^2}\right)^{3/2}}\rightarrow0 $$as $n\rightarrow\infty$ compare to $(\sum_{k=1}^n\frac{1}{k})^{-1/2}$.  We have $(S_n-A_n)/s_n\overset{d}{\rightarrow}N(0,1)$ where $$\begin{array}{rl}A_n&=\sum_{k=1}^n\alpha_{nk}=\sum_{k=1}^n\frac{1}{n-k+1}=\sum_{k=1}^n\frac{1}{k}; \\
s_n&=\sigma_n=\left(\sum_{k=1}^n\frac{1}{k}-\sum_{k=1}^n\frac{1}{k^2}\right)^{1/2}.\end{array}$$