# Sums to Products and Products to Sums

1. Nov 23, 2005

### benorin

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 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

$$\rho_{n} = \rho_{0} + \sum_{k=1}^{n} \left( \rho_{k} - \rho_{k-1} \right) = a_{0} + \sum_{k=1}^{n} \rho_{k-1} \left( a_{k} - 1 \right) = a_{0} + \sum_{k=1}^{n} a_{0}a_{1}\cdot\cdot\cdot a_{k-1} \left( a_{k} - 1 \right)$$

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 a very excellent text.

Last edited: Nov 23, 2005
2. Nov 24, 2005

### benorin

Oops, typo: that last tex line should read

$$\prod_{k=0}^{\infty} a_k = a_{0} + \sum_{k=1}^{\infty} a_{0}a_{1}\cdot\cdot\cdot a_{k-1} \left( a_{k} - 1 \right)$$

3. Jun 23, 2011

### Planck37

This is great work. Can someone prove this though or provide a link to a source please, though?

4. Jun 24, 2011

### Planck37

Your edit answers my question, thanks. "Theory and Applications of Infinite Series" by K. Knoppz.