Lyapunov CLT for Bernoulli independent random variables $X_n$ such that $P(X_n=1)\to0$ 
Let $\{X_n:n\ge1\}$ be a sequence of independent Bernoulli random variables with 
  $$P\{X_k=1\}=1-P\{X_k=0\}=\frac{1}{k}.$$
  Set
  $$S_n=\sum^{n}_{k=1}(X_k-\frac{1}{k}), \ B_n^2=\sum^{n}_{k=1}\frac{k-1}{k^2}$$
  Show that $\frac{S_n}{B_n}$ converges in distribution to the standard normal variable $Z$ as $n$ tends to infinity.

My attempt is to use the Lyapunov CLT, therefore we need to show there exists a $\delta>0$ such that,
$$\lim_{n\rightarrow \infty}\frac{1}{B_n^2}\sum_{k=1}^{n}E[|X_k-\frac{1}{k}|^{2+\delta}]=0. \ \ \ \ (1)$$
First I assume $\delta=1$ and look at the term $\sum_{k=1}^{n}E[|X_k-\frac{1}{k}|^{2+\delta}]$ we can see,
$$\sum_{k=1}^{n}E[(X_k-\frac{1}{k})^{3}]\le\sum_{k=1}^{n}E[|X_k-\frac{1}{k}|^{3}]\le\sum_{k=1}^{n}E[(X_k+\frac{1}{k})^{3}]$$.
I want to show the Lyapunov CLT holds for both $\sum_{k=1}^{n}E[(X_k-\frac{1}{k})^{3}]$ and $\sum_{k=1}^{n}E[(X_k+\frac{1}{k})^{3}]$ which would imply it holds for $\sum_{k=1}^{n}E[|X_k-\frac{1}{k}|^{3}]$.  
I found, 
$\sum_{k=1}^{n}E[(X_k+\frac{1}{k})^{3}]=\sum_{k=1}^{n} (\frac{1}{k}+\frac{3}{k^2}+\frac{4}{k^3})$ 
and
$\sum_{k=1}^{n}E[(X_k-\frac{1}{k})^{3}]=\sum_{k=1}^{n} (\frac{1}{k}-\frac{3}{k^2}+\frac{2}{k^3})$.
Now the problems I am having is with the $B_n^3$ term which is,
$B_n^3=(\sum^{n}_{k=1}\frac{k-1}{k^2})\sqrt{\sum^{n}_{k=1}\frac{k-1}{k^2}}$. This sum looks really ugly to me and I not sure if I can be simplified and even if I could I would still have to show (1) goes to zero. 
I was also looking at the Lindeberg condition to solve it but again I have to show for any $\epsilon>0$,
$$\lim_{n\rightarrow\infty}\frac{1}{\sum^{n}_{k=1}\frac{k-1}{k^2}}\sum^{n}_{k=1}E[X^2_k1_{\{ \frac{|X_k-\frac{1}{k}|}{\sqrt{\sum^{n}_{k=1}\frac{k-1}{k^2}}}>\epsilon\}}]=0.$$
Which I dont really understand especially the indicator function part of the expected value term. Any help or corrections would be appreciated.
 A: The Lyapunov condition holds for $\delta=1$ because
$$
\sum_{k=1}^{n}E|X_k-k^{-1}|^{3}=\sum_{k=1}^{n} \left(\frac{1}{k}-\frac{3}{k^2}+\frac{4}{k^3}-\frac{2}{k^4}\right)\le \sum_{k=1}^n\frac{1}{k}+C_1
$$
and
$$
B_n^2=\sum_{k=1}^n\left(\frac{1}{k}-\frac{1}{k^2}\right)\ge \sum_{k=1}^n\frac{1}{k}-C_2
$$
for some constants $C_1$ and $C_2$. Then (for $n\ge 3$ s.t. $\sum_{k\le n} k^{-1}>C_2$)
$$
B_n^{-3}\sum_{k=1}^{n}E|X_k-k^{-1}|^{3}\le \frac{\sum_{k\le n}k^{-1}+C_1}{\left(\sum_{k\le n}k^{-1}-C_2\right)^{3/2}}\to 0 \text{ as } n\to\infty.
$$

To see that the limit holds, you may use L'Hôpital's rule and the fact that $\sum_{k=1}^n k^{-1}=\psi(n+1)+\gamma$ and that $\psi(n)$ is increasing in $n$. Then
$$
\lim_{n\to\infty}\frac{\psi(n+1)+\gamma+C_1}{\left(\psi(n+1)+\gamma-C_2\right)^{3/2}}=\lim_{n\to\infty}\frac{2}{3}\frac{\psi^1(n+1)}{\psi^1(n+1)\left(\psi(n+1)+\gamma-C_2\right)^{1/2}}=0.
$$
A: This is more a comment
than an answer.
$\begin{array}\\
B_n^2
&=\sum^{n}_{k=1}\frac{k-1}{k^2}\\
&=\sum^{n}_{k=1}\frac{k}{k^2}-\sum^{n}_{k=1}\frac{1}{k^2}\\
&=\sum^{n}_{k=1}\frac{1}{k}-\sum^{n}_{k=1}\frac{1}{k^2}\\
&=\ln(n)+\gamma-\zeta(2)+O(\frac1{n})\\
&=\ln(n)+C+O(\frac1{n})
\qquad\text{where }C = \gamma-\zeta(2)\\
\text{so that}\\
B_n
&=\sqrt{\ln(n)}\sqrt{1+\frac{C}{\ln(n)}+O(\frac1{n \ln(n)}}\\
&=\sqrt{\ln(n)}(1+\frac{C}{2\ln(n)}+O(\frac1{\ln^2(n)})\\
&=\sqrt{\ln(n)}+\frac{C}{2\sqrt{\ln(n)}}+O(\frac1{\ln^{3/2}(n)})
\end{array}
$
