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Suppose that $I_i \sim Bern(\frac{1}{i})$. That is, $E(I_i) = \frac{1}{i}$ and $Var(I_i) = \frac{1}{i}-\frac{1}{i^2}$. Then, I want to show the following result about convergence in distribution.

$$ \frac{\sum_{i=1}^{n}I_i - \log n}{\sqrt{\log n}} \to_{D} N(0,1) $$

This problem is originally from Probability by Alan Gut and it describes the answer as:

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However, when he check's Lyapunov's condition, all he is doing is proving

$$ \frac{\sum_{i=1}^{n}(X_i-\frac{1}{i})}{\sqrt{\sum_{i=1}^{n}(\frac{1}{i}-\frac{1}{i^2})}} \to_{D} N(0,1) $$

However, he also tells us that $\gamma + \log n - \sum_{i=1}^{n}\frac{1}{i} \to 0$ as $n \to \infty$ where $\gamma \approx 0.577$ and

$$ \sum_{i=1}^{\infty}\frac{1}{i^2} = \frac{\pi^2}{6} $$

With these, how can I get the final form? By slutsky's?

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    Indeed, Slutsky's theorem is enough for finalizing the proof. – 2017-01-11
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    @SangchulLee I cannot get the final form, the information regarding the constants are all just simple limits, like $\lim_{n \to \infty}X_n = X$. Don't I need information about converging in distribution for the constants? – 2017-01-11
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    Let $(a_n)$ be a sequence of constants. Regarding them as constant random variables, convergence of $(a_n)$ as numbers is equivalent to a.s. convergence, which reduces to convergence in distribution. – 2017-01-11
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    @SangchulLee Thanks! – 2017-01-11
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    Slutsky is off-topic. The fact is that if $\frac{S_n-a_n}{b_n}$ converges in distribution to $N(0,1)$ with $a_n=\log n+O(1)$ and $b_n=\sqrt{\log n}+O(1)$ then $\frac{S_n-\log n}{\sqrt{\log n}}$ also converges in distribution to $N(0,1)$. Can you prove this in full generality? – 2017-01-11
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    @Did My confusion is how to incorporate the $\log n$ part, how would we do it? – 2017-01-11
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    I am afraid you will have to be **much more** specific and detailed about your "confusion"... – 2017-01-12
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    @Did I can prove $\frac{\sum_{i=1}^{n}(X_i-\frac{1}{i})}{\sqrt{\sum_{i=1}^{n}(\frac{1}{i}-\frac{1}{i^2})}} \to_{D} N(0,1)$, however, nowhere in here does the term $\log n$ appear. How can I go from this to what you have, which DOES have $\log n$? – 2017-01-12
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    $$\sum_{i=1}^n\left(X_i-\frac1i\right)=\sum_{i=1}^nX_i-\sum_{i=1}^n\frac1i\qquad\sum_{i=1}^n\left(\frac1i-\frac1{i^2}\right)=\sum_{i=1}^n\frac1i-\sum_{i=1}^n\frac1{i^2}$$ $$\sum_{i=1}^n\frac1i=\log n+O(1)\qquad\sum_{i=1}^n\frac1{i^2}=O(1)$$ – 2017-01-12
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    @Did Thanks, I understand that $O(1)$ means its some constants. How do we get rid of them or is it jus that they just disappear asymptotically? – 2017-01-12
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    Sorry but what do you think $Y_n/\sqrt{\log n}$ might converge to in distribution, if $|Y_n|\leqslant C$ almost surely and for every $n$, for some finite $C$? – 2017-01-13

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