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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
\sum_{k=0}^{\infty} a_k
Let S_n[/tex] denote the <i>n</i><sup>th</sup> partial sum, viz.<br /> <br /> S_n=\sum_{k=0}^{n} a_k<br /> <br /> so that, if S_{n}\neq 0,\forall n\in\mathbb{N} , then<br /> <br /> 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}}<br /> <br /> which is a pretty basic telescoping product, and it will simplify upon noticing that S_{k}= a_{k} + S_{k-1}, and that S_{0}= a_{0}, whence<br /> <br /> 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)<br /> <br /> and hence, taking the limit as n\rightarrow \infty, we have <br /> <br /> \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)<br /> <br /> now you can convert an infinite series to an infinite product. <br /> <br /> So the <i>vice-versa</i> part goes like this:<br /> <br /> Suppose the given product is<br /> <br /> \prod_{k=0}^{\infty} a_k<br /> <br /> Let \rho _n[/tex] denote the <i>n</i><sup>th</sup> partial product, viz.<br /> <br /> \rho_{n}=\prod_{k=0}^{n} a_k<br /> <br /> so that, if \rho_{n}\neq 0,\forall n\in\mathbb{N} , then<br /> <br /> \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)<br /> <br /> which is an extemely basic telescoping sum, and it will simplify upon noticing that \rho_{k}= a_{k} \rho_{k-1}, and that \rho_{0}= a_{0}, whence<br /> <br /> \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)<br /> <br /> and hence, taking the limit as n\rightarrow \infty, we have <br /> <br /> \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)<br /> <br /> and now you can convert an infinite product to an infinite series.<br /> <br /> So, go on, have fun with it... <br /> <br /> P.S. I swipped this technique from Theroy and Applications of Infinite Series by K. Knopp <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f609.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":wink:" title="Wink :wink:" data-smilie="2"data-shortname=":wink:" /> a very excellent text.
Suppose the given series is
\sum_{k=0}^{\infty} a_k
Let S_n[/tex] denote the <i>n</i><sup>th</sup> partial sum, viz.<br /> <br /> S_n=\sum_{k=0}^{n} a_k<br /> <br /> so that, if S_{n}\neq 0,\forall n\in\mathbb{N} , then<br /> <br /> 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}}<br /> <br /> which is a pretty basic telescoping product, and it will simplify upon noticing that S_{k}= a_{k} + S_{k-1}, and that S_{0}= a_{0}, whence<br /> <br /> 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)<br /> <br /> and hence, taking the limit as n\rightarrow \infty, we have <br /> <br /> \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)<br /> <br /> now you can convert an infinite series to an infinite product. <br /> <br /> So the <i>vice-versa</i> part goes like this:<br /> <br /> Suppose the given product is<br /> <br /> \prod_{k=0}^{\infty} a_k<br /> <br /> Let \rho _n[/tex] denote the <i>n</i><sup>th</sup> partial product, viz.<br /> <br /> \rho_{n}=\prod_{k=0}^{n} a_k<br /> <br /> so that, if \rho_{n}\neq 0,\forall n\in\mathbb{N} , then<br /> <br /> \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)<br /> <br /> which is an extemely basic telescoping sum, and it will simplify upon noticing that \rho_{k}= a_{k} \rho_{k-1}, and that \rho_{0}= a_{0}, whence<br /> <br /> \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)<br /> <br /> and hence, taking the limit as n\rightarrow \infty, we have <br /> <br /> \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)<br /> <br /> and now you can convert an infinite product to an infinite series.<br /> <br /> So, go on, have fun with it... <br /> <br /> P.S. I swipped this technique from Theroy and Applications of Infinite Series by K. Knopp <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f609.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":wink:" title="Wink :wink:" data-smilie="2"data-shortname=":wink:" /> a very excellent text.
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