# How to show $\frac{\sum_{i=1}^{n}X_i - \log n}{\sqrt{\log n}} \to_{D} N(0,1)$ for random variables $X_1, \ldots, X_n$ independent?

I currently have that for $X_1, \ldots, X_n$ independent Bernoulli random variables with parameter $\frac{1}{i}$ random variables,

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

This fact I established by Lyapunov's.

How can I show that if $\sum_{i=1}^n\frac1i=\log n+O(1)\qquad \ \text{and} \ \ \ \sum_{i=1}^n\frac1{i^2}=O(1)$,

Then:

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

?

Since the sum from 1 to n $1/i$ = ln n +O(1) you can substitute ln n for the sum $1/i$ in the numerator of your result and since sum $1/i^2 = O(1)$. sum $(1/i -1/i^2)=$ln n+$O(1)$ and ln n can be used to replace the quantity under the square root sign in the denominator.