This discussion is that of converting infinite series to infinite products and(adsbygoogle = window.adsbygoogle || []).push({}); vice-versain 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 then^{th}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 thevice-versapart goes like this:

Suppose the given product is

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

Let [itex]\rho _n[/tex] denote then^{th}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 a very excellent text.

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# Sums to Products and Products to Sums

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