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Math Help - A Series!

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

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

    Now: <br />
1 + \sum\limits_{k = 1}^\infty  {\tfrac{2}<br />
{{p^{k \cdot s} }}}  = 1 + \tfrac{2}<br />
{{p^s  - 1}} = \tfrac{{p^s  + 1}}<br />
{{p^s  - 1}} = \tfrac{{1 + \tfrac{1}<br />
{{p^s }}}}<br />
{{1 - \tfrac{1}<br />
{{p^s }}}}<br />

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

    Hence: <br />
\sum\limits_{n = 1}^\infty  {\tfrac{{2^{\nu \left( n \right)} }}<br />
{{n^s }}}  = \frac{{\zeta ^2 \left( s \right)}}<br />
{{\zeta \left( {2s} \right)}}<br />

    Set s=2 to get: <br />
\sum\limits_{n = 1}^\infty  {\tfrac{{2^{\nu \left( n \right)} }}<br />
{{n^2 }}}  = \frac{5}<br />
{2}<br />
since <br />
\zeta \left( 2 \right) = \tfrac{{\pi ^2 }}<br />
{6}{\text{ and }}\zeta \left( 4 \right) = \tfrac{{\pi ^4 }}<br />
{{90}}<br />
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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 <br />
2^{\nu \left( n \right)}<br />
is multiplicative ( i.e. given  <br />
a,b \in \mathbb{Z}^ + /\left( {a,b} \right) = 1<br />
we have <br />
2^{\nu \left( {a \cdot b} \right)} = 2^{\nu \left( a \right)} \cdot 2^{\nu\left( b \right)}<br />
)

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

    Now: <br />
1 + \sum\limits_{k = 1}^\infty {\tfrac{2}<br />
{{p^{k \cdot s} }}} = 1 + \tfrac{2}<br />
{{p^s - 1}} = \tfrac{{p^s + 1}}<br />
{{p^s - 1}} = \tfrac{{1 + \tfrac{1}<br />
{{p^s }}}}<br />
{{1 - \tfrac{1}<br />
{{p^s }}}}<br />

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

    Hence: <br />
\sum\limits_{n = 1}^\infty {\tfrac{{2^{\nu \left( n \right)} }}<br />
{{n^s }}} = \frac{{\zeta ^2 \left( s \right)}}<br />
{{\zeta \left( {2s} \right)}}<br />

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