How to show that $\frac1{\sqrt{\log n}}\left(\sum\limits_{i=1}^nI_i - \log n\right)$ converges to $N(0,1)$ for $I_i$ Bernoulli of parameter $\frac1i$?

Question

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:

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?