benorin
Nov23-05, 10:43 AM
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 nth partial sum, viz.
S_n=\sum_{k=0}^{n} a_k
so that, if [itex]S_{n}\neq 0,\forall n\in\mathbb{N} , then
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}}
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
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)
and hence, taking the limit as n\rightarrow \infty, we have
\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)
now you can convert an infinite series to an infinite product.
So the vice-versa part goes like this:
Suppose the given product is
\prod_{k=0}^{\infty} a_k
Let \rho _n[/tex] denote the nth partial product, viz.
\rho_{n}=\prod_{k=0}^{n} a_k
so that, if [itex]\rho_{n}\neq 0,\forall n\in\mathbb{N} , then
\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)
which is a pretty basic telescoping sum, and it will simplify upon noticing that \rho_{k}= a_{k} \rho_{k-1}, and that \rho_{0}= a_{0}, whence
\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)
and hence, taking the limit as n\rightarrow \infty, we have
\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)
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.
Edit: crap, meant to post this in the Calculus & Analysis forum, please delete this that I not be guilty of double posting...
Suppose the given series is
\sum_{k=0}^{\infty} a_k
Let S_n[/tex] denote the nth partial sum, viz.
S_n=\sum_{k=0}^{n} a_k
so that, if [itex]S_{n}\neq 0,\forall n\in\mathbb{N} , then
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}}
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
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)
and hence, taking the limit as n\rightarrow \infty, we have
\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)
now you can convert an infinite series to an infinite product.
So the vice-versa part goes like this:
Suppose the given product is
\prod_{k=0}^{\infty} a_k
Let \rho _n[/tex] denote the nth partial product, viz.
\rho_{n}=\prod_{k=0}^{n} a_k
so that, if [itex]\rho_{n}\neq 0,\forall n\in\mathbb{N} , then
\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)
which is a pretty basic telescoping sum, and it will simplify upon noticing that \rho_{k}= a_{k} \rho_{k-1}, and that \rho_{0}= a_{0}, whence
\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)
and hence, taking the limit as n\rightarrow \infty, we have
\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)
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.
Edit: crap, meant to post this in the Calculus & Analysis forum, please delete this that I not be guilty of double posting...