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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.

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  • $\begingroup$ Sorry but what exactly is preventing you to, first compute the exact value of each $E(|X_k-1/k|^3)$, then estimate the sums $\sum\limits_{k=1}^nE(|X_k-1/k|^3)$ when $n\to\infty$? Something escapes me here... $\endgroup$
    – Did
    Commented Mar 27, 2016 at 9:05

2 Answers 2

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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

$$ \lim_{n\to\infty}\frac{\psi(n+1)+\gamma+C_1}{\left(\psi(n+1)+\gamma-C_2\right)^{3/2}}=\lim_{n\to\infty}\frac{2}{3}\frac{\psi^1(n+1)}{\psi^1(n+1)\left(\psi(n+1)+\gamma-C_2\right)^{1/2}}=0. $$

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    $\begingroup$ Sometimes one feels that too much knowledge can be a trouble... Why not simply note that $\sum\limits_{k=1}^{n}E|X_k-k^{-1}|^{3}=\log n+O(1)$ and $B_n^2=\log n+O(1)$, and conclude directly from these? Better to summon L'Hôpital and the digamma function only when they are needed (which happens to be almost never, but this is another topic). $\endgroup$
    – Did
    Commented Mar 27, 2016 at 9:12
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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} $

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