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Sums to Products and Products to Sums

  1. Nov 23, 2005 #1

    benorin

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    This discussion is that of converting infinite series to infinite products and vice-versa in hopes of, say, ending the shortage of infinite product tables.

    Suppose the given series is

    [tex]\sum_{k=0}^{\infty} a_k[/tex]

    Let [itex]S_n[/tex] denote the nth partial sum, viz.

    [tex]S_n=\sum_{k=0}^{n} a_k[/tex]

    so that, if [itex]S_{n}\neq 0,\forall n\in\mathbb{N}[/itex] , then

    [tex]S_n=S_{0} \frac{S_{1}}{S_{0}} \frac{S_{2}}{S_{1}} \cdot\cdot\cdot \frac{S_{n}}{S_{n-1}} = S_{0} \prod_{k=1}^{n} \frac{S_{k}}{S_{k-1}}[/tex]

    which is a pretty basic telescoping product, and it will simplify upon noticing that [itex]S_{k}= a_{k} + S_{k-1}[/itex], and that [itex]S_{0}= a_{0}[/itex], whence

    [tex]S_n= S_{0} \prod_{k=1}^{n} \frac{S_{k}}{S_{k-1}} = a_{0} \prod_{k=1}^{n} \left( 1+ \frac{a_{k}}{S_{k-1}} \right) = a_{0} \prod_{k=1}^{n} \left( 1+ \frac{a_{k}}{a_{0}+a_{1}+\cdot\cdot\cdot + a_{k-1}} \right) [/tex]

    and hence, taking the limit as [itex] n\rightarrow \infty[/itex], we have

    [tex]\sum_{k=0}^{\infty} a_k = a_{0} \prod_{k=1}^{\infty} \left( 1+ \frac{a_{k}}{a_{0}+a_{1}+\cdot\cdot\cdot + a_{k-1}} \right) [/tex]

    now you can convert an infinite series to an infinite product.

    So the vice-versa part goes like this:

    Suppose the given product is

    [tex]\prod_{k=0}^{\infty} a_k[/tex]

    Let [itex]\rho _n[/tex] denote the nth partial product, viz.

    [tex]\rho_{n}=\prod_{k=0}^{n} a_k[/tex]

    so that, if [itex]\rho_{n}\neq 0,\forall n\in\mathbb{N}[/itex] , then

    [tex]\rho_{n} = \rho_{0} + \left( \rho_{1} - \rho_{0} \right) + \left( \rho_{2} - \rho_{1} \right) + \cdot\cdot\cdot + \left( \rho_{n} - \rho_{n-1} \right) = \rho_{0} + \sum_{k=1}^{n} \left( \rho_{k} - \rho_{k-1} \right) [/tex]

    which is an extemely basic telescoping sum, and it will simplify upon noticing that [itex]\rho_{k}= a_{k} \rho_{k-1}[/itex], and that [itex] \rho_{0}= a_{0}[/itex], whence

    [tex]\rho_{n} = \rho_{0} + \sum_{k=1}^{n} \left( \rho_{k} - \rho_{k-1} \right) = a_{0} + \sum_{k=1}^{n} \rho_{k-1} \left( a_{k} - 1 \right) = a_{0} + \sum_{k=1}^{n} a_{0}a_{1}\cdot\cdot\cdot a_{k-1} \left( a_{k} - 1 \right) [/tex]

    and hence, taking the limit as [itex] n\rightarrow \infty[/itex], we have

    [tex]\prod_{k=0}^{\infty} a_k = a_{0} + \sum_{k=1}^{n} a_{0}a_{1}\cdot\cdot\cdot a_{k-1} \left( a_{k} - 1 \right) [/tex]

    and now you can convert an infinite product to an infinite series.

    So, go on, have fun with it...

    P.S. I swipped this technique from Theroy and Applications of Infinite Series by K. Knopp :wink: a very excellent text.
     
    Last edited: Nov 23, 2005
  2. jcsd
  3. Nov 24, 2005 #2

    benorin

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    Oops, typo: that last tex line should read

    [tex]\prod_{k=0}^{\infty} a_k = a_{0} + \sum_{k=1}^{\infty} a_{0}a_{1}\cdot\cdot\cdot a_{k-1} \left( a_{k} - 1 \right) [/tex]
     
  4. Jun 23, 2011 #3
    This is great work. Can someone prove this though or provide a link to a source please, though?
     
  5. Jun 24, 2011 #4
    Your edit answers my question, thanks. "Theory and Applications of Infinite Series" by K. Knoppz.
     
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