For a set E of non-negative integers, let $Q(E)$ be the set of all integers $n = \Pi_p p^{\alpha (p)} $ where $\alpha (p) \in E$ for all primes p.
Show that, if $E = \{0, 1, 2, 5, 6, 7\},$ then
$$\sum_{n\in Q(E)} \frac {1}{n^s} =\frac{\zeta(s)\zeta(5s)}{ \zeta (3s) \zeta (10s )}$$
I really don't understand this question; for one $\alpha (p)=?$ .
$$\sum_{n\in Q(E)} \frac {1}{n^s} $$ $$= \Pi_{p}(1+\frac{1}{n^s}+\frac{1}{n^{2s}}+\frac{1}{n^{5s}}+\frac{1}{n^{6s}}+\frac{1}{n^{7s}})$$ $$= \Pi_{p}\bigg[(1+\frac{1}{n^s}+\frac{1}{n^{2s}})\cdot (1+\frac{1}{n^{5s}})\biggr]$$ $$= \Pi_{p}\bigg[\bigg(\frac{1-\frac{1}{n^{3s}}}{1-\frac{1}{p^s}}\biggr)\cdot\bigg(\frac{1-\frac{1}{n^{10s}}}{1-\frac{1}{n^{5s}}}\biggr)\biggr]$$ $$=\bigg[\bigg(\frac{\frac{1}{\zeta(3s)}}{\frac{1}{\zeta(s)}}\biggr)\cdot \bigg(\frac{\frac{1}{\zeta(10s)}}{\frac{1}{\zeta(5s)}}\biggr)\biggr]$$ $$=\frac{\zeta(s)\zeta(5s)}{ \zeta (3s) \zeta (10s )}$$