Convergence of $\sum_{n=2}^{\infty}\frac{1}{nQ(n)^{\lambda}}$ Let $Q(n)$ denote the largest prime factor of $n$. For which $\lambda > 0$ (perhaps all of them?) is it true that $\sum_{n=2}^{\infty}\frac{1}{n\cdot Q(n)^{\lambda}}$ converges?
My only ideas are bounds like $Q(n) \geq n^{1/\Omega(n)}$ where $\Omega(n)$ is the number of (not necessarily distinct) prime factors of $n$ and then show somehow that $\sum_{k=1}^{\infty}\sum_{\Omega(n) = k}\frac{1}{n^{1+\frac{\lambda}{k}}}$ converges - ideas for this or for something else?
Any help appreciated!
 A: (Partial answer) If $\lambda>1$, the series converges: $$\sum_n \frac 1 {nQ(n)^\lambda}=1+ \sum_{p~{\rm prime}}\sum_{m~{\rm s.t.~}Q(m)=p}\frac 1{mp^{\lambda}}<1+\sum_p\frac 1 {p^\lambda}<\sum_{n}\frac 1{n^\lambda}<\infty.$$
Note that for $m$ with $Q(m)=p,$ one has $$\frac 1 m=\frac 1{m'p}~{\rm with~}Q(m')\leq p$$ and $$\sum_{Q(m')\leq p}\frac 1{m'}=\prod_{q\leq p}\left(\frac 1{1-1/q}\right),{\rm ~with}~q~{\rm prime},$$ hence $$\sum_{Q(m)=p}\frac 1 m= \frac 2{2-1}\cdot \frac 3{3-1}\cdot \frac 5{5-1}\cdots \frac {p'}{p'-1}\cdot \frac p{p-1}\cdot \frac 1 p\leq 1,$$ where $p'$ is the prime right before $p$, and the inequality is strict if $p>2$.
A: Before starting, let us solidify some notation:
$$\mathbb{N}=\{1,2,...\}$$
$$\mathbb{N}_0=\{0,1,2,...\}$$
This would be true if the following conjecture held (It does hold, see EDIT below):
Conjecture: The sum
$$\sum_{z_i\in\mathbb{N}_0}\prod_{i=1}^k p_i^{-z_i}=O(\ln(k))$$
Basically, the sum over the inverse of the integers made up of the first $k$ primes (and $1$) grows like $\ln(k)$. For example
$$k=1:\ \sum_{i=0}^\infty \frac{1}{2^i}=2$$
$$k=2:\ \sum_{j=0}^\infty \sum_{i=0}^\infty \frac{1}{2^i 3^j}=3$$
Numerically this seems to be true and it makes sense because as $k$ goes to infinity this series looks more and more like the harmonic series (at least in a handwavy way). Additionally, the original question might still be true even if this conjecture is incorrect. If it turns out to be $O(\ln(k)\ln(\ln(k)))$ or something subpolynomial like that it would still work.
Proof: As suggested by @DesmondMiles we will split the sum into sections where each section has a largest prime $p_k$. Then
$$\sum_{n=2}^{\infty}\frac{1}{n\cdot Q(n)^{\lambda}}=\frac{1}{2^\lambda}\sum_{s\in \mathbb{N}}\frac{1}{2^{s}}+\frac{1}{3^\lambda}\sum_{z_1\in \mathbb{N}_0,s\in\mathbb{N}}\frac{1}{2^{z_1}3^{s}}+\frac{1}{5^\lambda}\sum_{z_i\in \mathbb{N}_0,s\in\mathbb{N}}\frac{1}{2^{z_1}3^{z_2}5^s}+...$$
$$=\frac{1}{2^{\lambda+1}}\sum_{s\in \mathbb{N}_0}\frac{1}{2^{s}}+\frac{1}{3^{\lambda+1}}\sum_{z_1\in \mathbb{N}_0,s\in\mathbb{N}_0}\frac{1}{2^{z_1}3^{s}}+\frac{1}{5^{\lambda+1}}\sum_{z_i\in \mathbb{N}_0,s\in\mathbb{N}_0}\frac{1}{2^{z_1}3^{z_2}5^s}+...$$
$$=\frac{1}{2^{\lambda+1}}\sum_{z_i\in \mathbb{N}_0}\frac{1}{2^{z_1}}+\frac{1}{3^{\lambda+1}}\sum_{z_i\in \mathbb{N}_0}\frac{1}{2^{z_1}3^{z_2}}+\frac{1}{5^{\lambda+1}}\sum_{z_i\in \mathbb{N}_0}\frac{1}{2^{z_1}3^{z_2}5^{z_3}}+...$$
$$=\sum_{k=1}^\infty\left[ \frac{1}{p_k^{\lambda+1}}\sum_{z_i\in\mathbb{N}_0}\prod_{i=1}^k p_i^{-z_i}\right]$$
Then with our conjecture this becomes
$$<C\sum_{k=2}^\infty\left[ \frac{1}{p_k^{\lambda+1}}\ln(k)\right]$$
for some constant $C$. Then utilizing the prime number theorem we have
$$\sim C\sum_{k=2}^\infty\left[ \frac{1}{(k\ln(k))^{\lambda+1}}\ln(k)\right]=C\sum_{k=2}^\infty\frac{1}{k^{\lambda+1}\ln(k)^{\lambda}}$$
which obviously converges since $\lambda>0$.

EDIT: So I put up the conjecture above as a question here. @subrosar answered and point me to Merten's Third Theorem. This theorem states that
$$e^{\gamma}\log(p_k)\sim \prod_{i=1}^k\left(\frac{1}{1-\frac{1}{p_i}}\right)=\prod_{i=1}^k \left(\sum_{j\in\mathbb{N}_0}^\infty p_i^{-j}\right)$$
Now notice that since the product is finite and the sum converges absolutely we are free to switch the two to get
$$=\sum_{z_i\in\mathbb{N}_0}\prod_{i=1}^k p_i^{-z_i}$$
Again calling on the prime number theorem we have
$$\sum_{z_i\in\mathbb{N}_0}\prod_{i=1}^k p_i^{-z_i}=\prod_{i=1}^k\left(\frac{1}{1-\frac{1}{p_i}}\right)\sim e^{\gamma}\ln(p_k)$$
$$\sim e^{\gamma}\ln(k\ln(k))=e^{\gamma}[\ln(k)+\ln(\ln(k))]\sim e^{\gamma}\ln(k)=O(\ln(k))$$
as desired.
A: Actually I found a way - if we partition according to the largest prime factor, then the contribution from the terms with $P = p_k$ (here $p_k$ is the $k$-th prime number) is (less than) $a_k = p_k^{-\lambda}\prod_{i=1}^k(1-1/p_i)^{-1}$. Finally, to show that $\sum_{k=1}^{\infty} a_k$ converges, one can use for example D'Alambert's ratio test, as well as (if really needed) asymptotics for $p_k$.
EDIT: D'Alambert doesn't help for the final step :( Any ideas?
EDIT: Ok, it's done now! Use the following weak form of Mertens' Third Estimate: $\prod_{p\leq x}(1-\frac{1}{p}) \leq C\log x$ (even $C(\log x)^d$ suffices) and conclude with the convergence of $\sum_{n=1}^{\infty} (\log n)^d/n^{1+\lambda}$ for $\lambda > 0$.
P.S. just observed after writing this, that QC_QAOA posted his conjecture in another thread and the answer to it there was exactly with Mertens' Third Estimate.
