Mean value of $\Omega(n)$ is $\log\log n$ This question is from my number theory assignment and I was unable to solve it. I have been following Dekonick and Luca.

Let $\Omega(n)$ be the number of prime divisors of n, counted with multiplicity ie the number of prime powers dividing n, for instance $\Omega(12) = 3$. Show that mean value of $\Omega(n)$ is $\log\log n$: $\frac{1}{N}\sum_{n\leq N} \Omega(n) = \log\log N+O(1)$.

Attempt: I thought of using Abel's Identity : $\sum_{n\leq N} f(n) = f(N)[N]- \int_{1}^x [t] f'(t)\,dt$.
Here $f(n) = f(p_1^{x_1} \cdots p_r^{x_r}) = x_1 + \dots + x_r$
So, I got $\sum_{n\leq N} f(n) = (x_1 + \dots + x_r)[N]- \int_{1}^{x} [t] (x_1+\dots+x_r) \log t\,dt$. Dividing by $N$, $((x_1 + \dots + x_r)[N])/N =O(1)$, and the other term after simplifying I got as $$(x_1+\dots+x_r)(\log N -1-(N\log N)/2+N/4+1/N(3/4))$$ which is not equal to what has to be proved.
So, please help.

By definition, we know
\begin{aligned}
0\le\sum_{n\le N}[\Omega(n)-\omega(n)]
&=\sum_{n\le N}\sum_{p^a\|n,a>1}a\le\sum_{a\ge2}a\sum_{p\le N}\sum_{\substack{n\le N\\p^a|n}}1 \\
&\le\sum_{a\ge2}a\sum_{p\le N}{N\over p^a}\le N\sum_{p\le N}\sum_{a\ge2}{a\over p^a}
\end{aligned}
To evaluate the rightmost sum, we recall the properties of geometric series. That is, for $|z|<1$ there is
$$
{1\over1-z}=1+z+\sum_{r\ge2}z^r
$$
Differentiating on both side, we have
$$
{1\over(1-z)^2}=1+\sum_{r\ge2}rz^{r-1}
$$
Multiplying $z$ on both side gives
$$
\sum_{r\ge2}rz^r={z-z(z^2-2z+1)\over(1-z)^2}={z^2(2-z)\over(1-z)^2}
$$
This indicates that
$$
\sum_{p\le N}\sum_{a\ge2}{a\over p^a}=\sum_{p\le N}{p^{-2}(2-p^{-1})\over(1-p^{-1})^2}\le\sum_{p\le N}{2\over(p-1)^2}
$$
Since the right most sum is convergent, we conclude
$$
\sum_{n\le N}[\Omega(n)-\omega(n)]=\mathcal O(N)
$$
which indicates that
$$
\sum_{n\le N}\Omega(n)=\sum_{n\le N}\omega(n)+\mathcal O(N)\tag1
$$
which converts our problem into a much more convenient one:
\begin{aligned}
\sum_{n\le N}\omega(n)
&=\sum_{n\le N}\sum_{p|n}1=\sum_{p\le N}\sum_{\substack{n\le N\\p|n}}1 \\
&=\sum_{p\le N}\left\lfloor\frac Np\right\rfloor=N\sum_{p\le N}\frac1p+\mathcal O(N)
\end{aligned}
To continue, we summon Mertens' formula (a superior version is presented as Theorem 427 in Hardy & Wright):
$$
\sum_{p\le N}\frac1p=\log\log N+\mathcal O(1)
$$
so that
$$
\sum_{n\le N}\omega(n)=N\log\log N+\mathcal O(N)\tag2
$$
Combining (1) and (2) gives the desired result.
