evaluating an infinite product

**rn443** · October 9th 2009, 10:40 PM

How do I evaluate $\prod_{i=1}^\infty \frac{10^i - 1}{10^i} = .9\cdot .99 \cdot .99 \cdot \ldots$ ? Note that it's equivalent to solving the recurrence relation $p_{n+1} = p_n - \frac{p_n}{10^{n+1}}$ for $p_1 = \frac{9}{10}$ .

I know how to prove it's between .8 and 1, but I can't figure out an exact answer.

**Laurent** · October 10th 2009, 02:07 AM

Originally Posted by rn443

How do I evaluate $\prod_{i=1}^\infty \frac{10^i - 1}{10^i} = .9\cdot .99 \cdot .99 \cdot \ldots$ ? Note that it's equivalent to solving the recurrence relation $p_{n+1} = p_n - \frac{p_n}{10^{n+1}}$ for $p_1 = \frac{9}{10}$ .

I know how to prove it's between .8 and 1, but I can't figure out an exact answer.

I don't think there is a closed formula. There is a famous formula by Euler that goes as follow : $\prod_{n=1}^\infty (1-q^n) = \frac{1}{\sum_{n=0}^\infty p(n) q^n}$ where $p(n)$ is the number of partitions of $n$ . But no explicit formula is usually given; that says it all.

**Opalg** · October 10th 2009, 12:55 PM

Originally Posted by Laurent

I don't think there is a closed formula. There is a famous formula by Euler that goes as follow : $\prod_{n=1}^\infty (1-q^n) = \frac{1}{\sum_{n=0}^\infty p(n) q^n}$ where $p(n)$ is the number of partitions of $n$ . But no explicit formula is usually given; that says it all.

There is some more information about this function here. That link gives the expansion of $\prod_{n=1}^\infty (1-x^n)$ as $1 - x - x^2 + x^5 + x^7 - x^{12} - x^{15} + x^{22} + x^{26} - x^{35} - x^{40} + x^{51} + \ldots$ . It's not stated in the link, but it looks very much as though the nonzero coefficients are all $\pm1$ (with a pattern of two –1s followed by two +1s). If so then $\prod_{n=1}^\infty (1-10^{-n}) = 0.8900101 - 10^{-12} - 10^{-15}$ with an error of less than $10^{-21}$ .

In fact, that pattern of pairs of ±1s looks extremely regular, with the k'th pair being the coefficients of $x^{p_k}$ and $x^{q_k}$ , where $p_k=\tfrac12k(3k-1)$ and $q_k=\tfrac12k(3k+1)$ . Again, that's just based on observations from the list of the first 200 coefficients in the Encyclopedia of Integer Sequences. I don't have any proof for it.

**Laurent** · October 10th 2009, 01:09 PM

Originally Posted by Opalg

There is some more information about this function here. That link gives the expansion of $\prod_{n=1}^\infty (1-x^n)$ as $1 - x - x^2 + x^5 + x^7 - x^{12} - x^{15} + x^{22} + x^{26} - x^{35} - x^{40} + x^{51} + \ldots$ . It's not stated in the link, but it looks very much as though the nonzero coefficients are all $\pm1$ (with a pattern of two –1s followed by two +1s). If so then $\prod_{n=1}^\infty (1-10^{-n}) = 0.8900101 - 10^{-12} - 10^{-15}$ with an error of less than $10^{-21}$ .

In fact, that pattern of pairs of ±1s looks extremely regular, with the k'th pair being the coefficients of $x^{p_k}$ and $x^{q_k}$ , where $p_k=\tfrac12k(3k-1)$ and $q_k=\tfrac12k(3k+1)$ . Again, that's just based on observations from the list of the first 200 coefficients in the Encyclopedia of Integer Sequences. I don't have any proof for it.

You seem to be correct. It is called Euler's pentagonal theorem and two proofs are given here.

**chisigma** · October 10th 2009, 09:03 PM

A reasonable alternative to the 'direct computation' of the 'infinite product' is the computation of its logarithm...

$L(x)= \ln \prod_{n=1}^{\infty} (1-x^{n}) = \sum_{n=1}^{\infty} \ln (1-x^{n})$ (1)

Using the standard Taylor expansion of $\ln (1-x^{n})$ we obtain...

$L(x)= -\sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{x^{kn}}{k} = \sum_{k=1}^{\infty}\frac {x^{k}}{k\cdot (x^{k}-1)}$ (2)

Both the 'infinite product' and the 'infinite sum' are convergent for $|x|<1$ . The computation using the first three hundred terms of the 'product' and the first three hundred terms of the 'sum' leads us to the same result approximated in twelve digits...

$\prod_{n=1}^{\infty} (1-.1^{n}) \approx .890010099999...$

Kind regards

$\chi$ $\sigma$

**aman_cc** · October 10th 2009, 11:19 PM

Originally Posted by chisigma

A reasonable alternative to the 'direct computation' of the 'infinite product' is the computation of its logarithm...

$L(x)= \ln \prod_{n=1}^{\infty} (1-x^{n}) = \sum_{n=1}^{\infty} \ln (1-x^{n})$ (1)

Using the standard Taylor expansion of $\ln (1-x^{n})$ we obtain...

$L(x)= -\sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{x^{kn}}{k} = \sum_{k=1}^{\infty}\frac {x^{k}}{k\cdot (x^{k}-1)}$ (2)

Both the 'infinite product' and the 'infinite sum' are convergent for $|x|<1$ . The computation using the first three hundred terms of the 'product' and the first three hundred terms of the 'sum' leads us to the same result approximated in twelve digits...

$\prod_{n=1}^{\infty} (1-.1^{n}) \approx .890010099999...$

Kind regards

$\chi$ $\sigma$

A few related question please -
1. Is there a easy way to show if the result is rational or irrational?
2. @rn443: Can you please share how you proved that the result is greater than 0.8?

Thanks

**Laurent** · October 11th 2009, 02:16 AM

Originally Posted by aman_cc

A few related question please -
1. Is there a easy way to show if the result is rational or irrational?

Yes.

Because of the formula

$\prod_{n=1}^\infty (1+q^n)=1+\sum_{n=1}^\infty (-1)^n (q^{\frac{n(3n-1)}{2}}+q^{\frac{n(3n+1)}{2}})$ ,

we have $\alpha=\prod_{n=1}^\infty \left(1+\frac{1}{10^n}\right) =$ $1.00001010000000000000010001\ldots - 0.11000000000100100\ldots$ (only ones and zeros).

Note as well that $0.111111\cdots = \frac{1}{9}$ . Therefore,

$\alpha + \frac{1}{9} = 1.0011212111101101111112111211\cdots$ .

There may be mistakes, but you see the idea: $\alpha+\frac{1}{9}$ has a decimal expansion with only 0,1,2's. And because the exponents in the first formula are quadratic (not linear), this expansion is not eventually periodic.

We conclude that $\alpha+\frac{1}{9}$ is irrational, and thus so is $\alpha$ .

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