1. ## A Series!

This is nice and it shouldn't be hard: Let $\displaystyle \nu(n)$ be the number of distinct prime factors of $\displaystyle n.$ We define $\displaystyle \nu(1)=0.$ Evaluate $\displaystyle S=\sum_{n=1}^{\infty} \frac{2^{\nu(n)}}{n^2}.$

2. It's fairly easy to see that $\displaystyle 2^{\nu \left( n \right)}$ is multiplicative ( i.e. given $\displaystyle a,b \in \mathbb{Z}^ + /\left( {a,b} \right) = 1$ we have $\displaystyle 2^{\nu \left( {a \cdot b} \right)} = 2^{\nu \left( a \right)} \cdot 2^{\nu\left( b \right)}$ )

Then: $\displaystyle \sum\limits_{n = 1}^\infty {\tfrac{{2^{\nu \left( n \right)} }} {{n^s }}} = \prod\limits_{p{\text{ prime}}} {\left( {\sum\limits_{k = 0}^\infty {\tfrac{{2^{\nu \left( {p^k } \right)} }} {{p^{k \cdot s} }}} } \right)} = \prod\limits_{p{\text{ prime}}} {\left( {1 + \sum\limits_{k = 1}^\infty {\tfrac{2} {{p^{k \cdot s} }}} } \right)}$

Now: $\displaystyle 1 + \sum\limits_{k = 1}^\infty {\tfrac{2} {{p^{k \cdot s} }}} = 1 + \tfrac{2} {{p^s - 1}} = \tfrac{{p^s + 1}} {{p^s - 1}} = \tfrac{{1 + \tfrac{1} {{p^s }}}} {{1 - \tfrac{1} {{p^s }}}}$

$\displaystyle \sum\limits_{n = 1}^\infty {\tfrac{{2^{\nu \left( n \right)} }} {{n^s }}} = \prod\limits_{p{\text{ prime}}} {\left( {\tfrac{{1 + \tfrac{1} {{p^s }}}} {{1 - \tfrac{1} {{p^s }}}}} \right)} = \underbrace {\prod\limits_{p{\text{ prime}}} {\left( {\tfrac{1} {{1 - \tfrac{1} {{p^s }}}}} \right)} }_{\zeta \left( s \right)} \cdot \prod\limits_{p{\text{ prime}}} {\left( {1 + \tfrac{1} {{p^s }}} \right)}$; $\displaystyle \prod\limits_{p{\text{ prime}}} {\left( {1 + \tfrac{1} {{p^s }}} \right)} = \frac{{\prod\limits_{p{\text{ prime}}} {\left( {1 - \tfrac{1} {{p^{2s} }}} \right)} }} {{\prod\limits_{p{\text{ prime}}} {\left( {1 - \tfrac{1} {{p^s }}} \right)} }} = \frac{{\zeta \left( s \right)}} {{\zeta \left( {2s} \right)}}$

Hence: $\displaystyle \sum\limits_{n = 1}^\infty {\tfrac{{2^{\nu \left( n \right)} }} {{n^s }}} = \frac{{\zeta ^2 \left( s \right)}} {{\zeta \left( {2s} \right)}}$

Set $\displaystyle s=2$ to get: $\displaystyle \sum\limits_{n = 1}^\infty {\tfrac{{2^{\nu \left( n \right)} }} {{n^2 }}} = \frac{5} {2}$ since $\displaystyle \zeta \left( 2 \right) = \tfrac{{\pi ^2 }} {6}{\text{ and }}\zeta \left( 4 \right) = \tfrac{{\pi ^4 }} {{90}}$

3. Originally Posted by PaulRS
It's fairly easy to see that $\displaystyle 2^{\nu \left( n \right)}$ is multiplicative ( i.e. given $\displaystyle a,b \in \mathbb{Z}^ + /\left( {a,b} \right) = 1$ we have $\displaystyle 2^{\nu \left( {a \cdot b} \right)} = 2^{\nu \left( a \right)} \cdot 2^{\nu\left( b \right)}$ )

Then: $\displaystyle \sum\limits_{n = 1}^\infty {\tfrac{{2^{\nu \left( n \right)} }} {{n^s }}} = \prod\limits_{p{\text{ prime}}} {\left( {\sum\limits_{k = 0}^\infty {\tfrac{{2^{\nu \left( {p^k } \right)} }} {{p^{k \cdot s} }}} } \right)} = \prod\limits_{p{\text{ prime}}} {\left( {1 + \sum\limits_{k = 1}^\infty {\tfrac{2} {{p^{k \cdot s} }}} } \right)}$

Now: $\displaystyle 1 + \sum\limits_{k = 1}^\infty {\tfrac{2} {{p^{k \cdot s} }}} = 1 + \tfrac{2} {{p^s - 1}} = \tfrac{{p^s + 1}} {{p^s - 1}} = \tfrac{{1 + \tfrac{1} {{p^s }}}} {{1 - \tfrac{1} {{p^s }}}}$

$\displaystyle \sum\limits_{n = 1}^\infty {\tfrac{{2^{\nu \left( n \right)} }} {{n^s }}} = \prod\limits_{p{\text{ prime}}} {\left( {\tfrac{{1 + \tfrac{1} {{p^s }}}} {{1 - \tfrac{1} {{p^s }}}}} \right)} = \underbrace {\prod\limits_{p{\text{ prime}}} {\left( {\tfrac{1} {{1 - \tfrac{1} {{p^s }}}}} \right)} }_{\zeta \left( s \right)} \cdot \prod\limits_{p{\text{ prime}}} {\left( {1 + \tfrac{1} {{p^s }}} \right)}$; $\displaystyle \prod\limits_{p{\text{ prime}}} {\left( {1 + \tfrac{1} {{p^s }}} \right)} = \frac{{\prod\limits_{p{\text{ prime}}} {\left( {1 - \tfrac{1} {{p^{2s} }}} \right)} }} {{\prod\limits_{p{\text{ prime}}} {\left( {1 - \tfrac{1} {{p^s }}} \right)} }} = \frac{{\zeta \left( s \right)}} {{\zeta \left( {2s} \right)}}$

Hence: $\displaystyle \sum\limits_{n = 1}^\infty {\tfrac{{2^{\nu \left( n \right)} }} {{n^s }}} = \frac{{\zeta ^2 \left( s \right)}} {{\zeta \left( {2s} \right)}}$

Set $\displaystyle s=2$ to get: $\displaystyle \sum\limits_{n = 1}^\infty {\tfrac{{2^{\nu \left( n \right)} }} {{n^2 }}} = \frac{5} {2}$ since $\displaystyle \zeta \left( 2 \right) = \tfrac{{\pi ^2 }} {6}{\text{ and }}\zeta \left( 4 \right) = \tfrac{{\pi ^4 }} {{90}}$
correct! well done!