Proof that the series $\sum_{n=2}^{\infty}\frac{[\Omega(n)]^\alpha}{n^2}$ converges

Question

Let's consider the series $$f(\alpha)=\sum_{n\gt1}\frac{[\,\Omega(n)\,]^\alpha}{n^2}$$ where $\Omega(n)$ denotes the number of prime factors of $n$ counted with their multiplicity and $\alpha\ge0$ is a real parameter.

It is known that $$f(0)=\frac{\pi^2}{6}-1$$ and by computational experiments results that $$f\Big(\frac 1 2\Big)=0.74587577\dots$$ $$f(1)=0.90748082\dots$$ $$f(2)=1.62036452\dots$$ How to prove that $f(\alpha)$ is finite (that is the given series converges) for any value of $\alpha$?

The trend of the function $\log f(\alpha)$ is shown in the following graph:

One could conjecture that $$f(\alpha)\sim e^{C\alpha}$$ with $C=\frac 5 2$ approximately.

How to prove this estimate?

Answer 1

The finiteness (and the asymptotics) follow by partial summation from the estimate that the summatory function of $\Omega(n)^a$ is $x (\log \log x)^a$ See The Wikipedia article for the latter.

Answer 2

Too long of a comment. How did you come up with your estimate? If I set $$\Omega(n)=\{n_1+...+n_i+... | n=p_1^{n_1} \cdots p_i^{n_i} \cdots\}$$ for a finite number of $n_i\neq 0$, then $\Omega(n)=k$ includes all the numbers of the form $n=2^k,3^k,5^k,...$ besides mixed numbers and so $$\sum_{n=2}^\infty \frac{(\Omega(n))^\alpha}{n^2} = \sum_{k=1}^\infty \sum_{\substack{n=2 \\ \Omega(n)=k}}^\infty \frac{k^\alpha}{n^2} \geq \sum_{k=1}^\infty\sum_{p\in\mathbb{P}} \frac{k^\alpha}{p^{2k}} \geq \sum_{k=1}^\infty \frac{k^\alpha}{2^{2k}} \\ \sim \int_0^\infty {\rm d}k \, k^\alpha 4^{-k} = \frac{\Gamma(\alpha+1)}{(\log 4)^{\alpha+1}} \sim \frac{\sqrt{2\pi \alpha}}{\log 4} \left(\frac{\alpha}{e\log 4}\right)^\alpha$$ which is eventually larger than your $\sim e^{5\alpha/2}$ asymptotics.