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!

(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 $p>2$.
