The normal order of $\nu(n)$ and its normal distribution I understand that for $\nu(n)$, the number of divisors of $n$, let me denote $\log_2 x=\log\log x$. Then $$\sum_{n\le x}\Big(\nu(n)-\log_2x\Big)^2=O(x\log_2x)$$
The book told me about the theorem (due to Erdos and Kac), for $S(x;\alpha,\beta)=\#\left\{ n\le x : \alpha\le\dfrac{\nu(n)-\log_2n}{\sqrt{\log_2(n)}}\le\beta\right\}:=\#S$, then $$\lim_{x\rightarrow\infty}\dfrac{S(x;\alpha,\beta)}{x}=\dfrac{1}{\sqrt{2\pi}}\int_{\alpha}^\beta e^{-t^2/2} dt$$
I want to prove that $\displaystyle\sum_{n\le x}\Big(\nu(n)-\log_2n\Big)^2\ll x\log_2 x$ by using the theorem stated above.
What I am confused is that if $n\in S$ then $\displaystyle \delta(n):=\left|\dfrac{\nu(n)-\log_2 n}{\sqrt{\log_2 (n)}}\right|\le \max(|\alpha|,|\beta|)$ and if we choose $|\alpha|,|\beta|\le \sqrt x$ then the statement I want follows for $n\in S$, but what about $n\not\in S, n\le x$? I confuse about the phrase "normally distributed". What does it mean exactly in this sense?
As I guess, if we choose $\alpha,\beta$ that $\beta-\alpha$ is wide enough then $\delta(n)$ that satisfies the condition in $S$ would be wide enough, too. Therefore, the number of terms left from the distribution (the term on the left and right part of normal distribution) would be small.
How can I add the left terms that I mentioned to cover all $n\le x$?
 A: I proved it by using the method that resembled Turan that
\begin{equation}
\begin{split}
\sum_{1<n\le x}\nu(n) & =x\log_2x+O(x) \\
\sum_{n\le x} \nu(n)^2&= x(\log_2x)^2+O(x\log_2x)
\end{split}
\end{equation}
and by partial summation
\begin{equation}
\begin{split}
\sum_{1<n\le x}(\log_2 n)^2 & =\sum_{1<n\le x}1\cdot(\log_2n)^2 = (\left[\,x\right]-1)\log^2_2x-\int_2^x (\left[\,t\right]-1)\cdot 2\log_2 t\cdot\left(\dfrac{1}{t\log t}\right) dt\\
&= x(\log_2 x)^2+O(\log^2_2 x)+O(x\log_2 x)= x(\log_2 x)^2+O(x\log_2 x)\\
\sum_{1<n\le x} \nu(n)\log_2 n & = \sum_{n\le x} \nu(n)\cdot \log_2 n=\Big(x\log_2 x+O(x)\Big)(\log_2 x)-\int_2^x (t\log_2 t+O(t))\left(\dfrac{1}{t\log t}\right) dt \\
& = x(\log_2 x)^2+O(x\log_2 x)
\end{split}
\end{equation}
Hence, we obtain
\begin{equation}
\begin{split}
\sum_{1<n\le x}(\nu(n)-\log_2 n)^2 & = \sum \nu(n)^2-2\sum\nu(n)\log_2n+\sum(\log_2 n)^2 = O(x\log_2 x)
\end{split}
\end{equation}
as desired. Therefore, almost all numbers $n$ have $\log_2 n$ numbers of prime factors as Hardy and Ramanujan stated (I find their theorem indeed elegant).
A: It is not possible in general to show that weak convergence implies convergence of moments. One needs some extra assumption, which in your example just happens to be the bound that you want to prove.
