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Thread: A Series!

  1. #1
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    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 $\displaystyle S=\sum_{n=1}^{\infty} \frac{2^{\nu(n)}}{n^2}.$
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  2. #2
    Super Member PaulRS's Avatar
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    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}}
    $
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  3. #3
    MHF Contributor

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    Quote Originally Posted by PaulRS View Post
    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!
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