# Math Help - A Series!

1. ## A Series!

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

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

Then: $
\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: $
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 }}}}
$

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

Set $s=2$ to get: $
\sum\limits_{n = 1}^\infty {\tfrac{{2^{\nu \left( n \right)} }}
{{n^2 }}} = \frac{5}
{2}
$
since $
\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 $
2^{\nu \left( n \right)}
$
is multiplicative ( i.e. given $
a,b \in \mathbb{Z}^ + /\left( {a,b} \right) = 1
$
we have $
2^{\nu \left( {a \cdot b} \right)} = 2^{\nu \left( a \right)} \cdot 2^{\nu\left( b \right)}
$
)

Then: $
\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: $
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 }}}}
$

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

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