- Thread starter
- #1

- Thread starter bincybn
- Start date

- Thread starter
- #1

- Jan 26, 2012

- 890

If that is what Mathematica thinks it sums to then you can be reasonably confident that there is no simple closed form for that sum in terms of elementary functions.View attachment 145 where M>=2. A close upper bound also will be useful(not like 1 as the upper bound). Thanks in advance.

This is also QPochhammer[1/M,1/M,inf]. Courtesy to mathematica.

I can think of a couple of methods of getting upper bounds, the first is to take logs then use the first term of the power series expansion of:

\(\displaystyle \log\left(1-\frac{1}{M^i}\right)<- \frac{1}{M^i} \)

so:

\(\displaystyle \sum_{i=1}^{\infty} \left(\log \left(1-\frac{1}{M^i} \right) \right) <- \sum_{i=1}^{\infty} \frac{1}{M^i}= -\; \frac{1}{M}\;\frac{1}{1-\frac{1}{M}}=\frac{1}{1-M} \)

Then exponentiating we get:

\( \displaystyle \prod_{i=1}^{\infty} \left(1-\frac{1}{M^i}\right) < e^{\frac{1}{1-M}} \)

A second way is to bound the infinite sum above by intergrals.

CB

- Thread starter
- #3

- Thread starter
- #4

I think, i got the ans for my own previous question. Can anyone pls verify it? All suggestions are always welcome.\(\displaystyle \prod_{i=1}^{\infty}\left(1-\frac{1}{M^{i}}\right)\) converges(Thanks to Mr. CaptainBlack for the upper bound) strictly above zero.

Consider a problem as follows: (I posted this in some other forum for some other purpose. Here I am repeating it for a different purpose. Hope that it won't violate the rules)

I have divided time into different slots, transmitting different coins(one in each slot) with probability of heads Bernoulli\(\displaystyle P_{k}=\frac{1}{M^{k}}\) where k is the slot index.

Then \(\displaystyle \prod_{i=1}^{\infty}\left(1-\frac{1}{M^{i}}\right)\) is the P (Head never happens) = 1-P (Head ever happens) =1-\(\displaystyle \sum_{i=1}^{\infty} \left\{ \prod_{j=1}^{i-1}\left(1-\frac{1}{M^{j}}\right)\right\} *\left(\frac{1}{M^{i}}\right)\)

\(\displaystyle \sum_{i=1}^{\infty} \left\{ \prod_{j=1}^{i-1}\left(1-\frac{1}{M^{j}}\right)\right\} *\left(\frac{1}{M^{i}}\right)\) < \(\displaystyle \sum_{i=1}^{\infty} \frac{1}{M^{i}}\) since \(\displaystyle \left\{ \prod_{j=1}^{i-1}\left(1-\frac{1}{M^{j}}\right)\right\}\) always less than 1 \(\displaystyle \forall M \geq2\) .

\(\displaystyle \Longrightarrow \sum_{i=1}^{\infty} \left\{ \prod_{j=1}^{i-1}\left(1-\frac{1}{M^{j}}\right)\right\} *\left(\frac{1}{M^{i}}\right)\) < \(\displaystyle \frac{1}{M-1} \leq 1\)(=1 for M=2)

Therefore \(\displaystyle \sum_{i=1}^{\infty} \left\{ \prod_{j=1}^{i-1}\left(1-\frac{1}{M^{j}}\right)\right\} *\left(\frac{1}{M^{i}}\right)\) < 1

and \(\displaystyle \prod_{i=1}^{\infty}\left(1-\frac{1}{M^{i}}\right)\) > 0.

Consider a problem as follows: (I posted this in some other forum for some other purpose. Here I am repeating it for a different purpose. Hope that it won't violate the rules)

I have divided time into different slots, transmitting different coins(one in each slot) with probability of heads Bernoulli\(\displaystyle P_{k}=\frac{1}{M^{k}}\) where k is the slot index.

Then \(\displaystyle \prod_{i=1}^{\infty}\left(1-\frac{1}{M^{i}}\right)\) is the P (Head never happens) = 1-P (Head ever happens) =1-\(\displaystyle \sum_{i=1}^{\infty} \left\{ \prod_{j=1}^{i-1}\left(1-\frac{1}{M^{j}}\right)\right\} *\left(\frac{1}{M^{i}}\right)\)

\(\displaystyle \sum_{i=1}^{\infty} \left\{ \prod_{j=1}^{i-1}\left(1-\frac{1}{M^{j}}\right)\right\} *\left(\frac{1}{M^{i}}\right)\) < \(\displaystyle \sum_{i=1}^{\infty} \frac{1}{M^{i}}\) since \(\displaystyle \left\{ \prod_{j=1}^{i-1}\left(1-\frac{1}{M^{j}}\right)\right\}\) always less than 1 \(\displaystyle \forall M \geq2\) .

\(\displaystyle \Longrightarrow \sum_{i=1}^{\infty} \left\{ \prod_{j=1}^{i-1}\left(1-\frac{1}{M^{j}}\right)\right\} *\left(\frac{1}{M^{i}}\right)\) < \(\displaystyle \frac{1}{M-1} \leq 1\)(=1 for M=2)

Therefore \(\displaystyle \sum_{i=1}^{\infty} \left\{ \prod_{j=1}^{i-1}\left(1-\frac{1}{M^{j}}\right)\right\} *\left(\frac{1}{M^{i}}\right)\) < 1

and \(\displaystyle \prod_{i=1}^{\infty}\left(1-\frac{1}{M^{i}}\right)\) > 0.

Last edited:

- Feb 13, 2012

- 1,704

The function...View attachment 145 where M>=2. A close upper bound also will be useful(not like 1 as the upper bound). Thanks in advance.

This is also QPochhammer[1/M,1/M,inf]. Courtesy to mathematica.

$\displaystyle \phi(z)=\prod _{n=1}^{\infty} (1-z^{n})$ (1)

... is know as 'Euler's Function'. An explicit elementary expression of (1) is not know so that we try the series expansion of its logarithm...

$\displaystyle \ln \phi(z)= \sum_{n=1}^{\infty} \ln (1-z^{n})= -\sum_{n=1}^{\infty}\ \sum_{k=1}^{\infty} \frac{z^n k}{k}= -\sum_{k=1}^{\infty} \frac{1}{k}\ \sum_{n=1}^{\infty} z^{n k}= - \sum_{k=1}^{\infty} \frac{z^{k}}{k\ (1-z^{k})}$ (2)

Any finite sum of (2) is an 'upper bound' of the function and if k increases then the 'upper bound' is more close to the function. Setting $z=\frac{1}{m}$ in (2) if You uses only the first term of the series You obtain...

$\displaystyle \prod _{n=1}^{\infty} (1-m^{-n}) < e^{- \frac{1}{m-1}}$ (3)

If You use two terms of the series expansion You obtain...

$\displaystyle \prod _{n=1}^{\infty} (1-m^{-n}) < e^{- \frac{1}{m-1}}\ e^{- \frac{1}{2\ (m^{2}-1)}}$ (4)

... and so one...

Kind regards

$\chi$ $\sigma$

- Moderator
- #6

- Feb 7, 2012

- 2,799

You can get it bounded away from 0 (for any $m>1$) like this. First, for $0<x<1$, $$ -\ln(1-x) = x + \tfrac{x^2}2 + \tfrac{x^3}3 + \ldots < x + x^2 + x^3 + \ldots = \tfrac x{1-x}.$$Can you tell anything about the lower bound? My doubt is whether it will converge to zero or not?

Then $\displaystyle -\sum_{n=1}^\infty\ln(1-m^{-n}) < \sum_{n=1}^\infty \tfrac1{m^n-1}.$ But $m^n-1 = (m-1)(m^{n-1} + \ldots + 1) > m^{n-1}(m-1)$, and therefore $$-\sum_{n=1}^\infty\ln(1-m^{-n}) < \sum_{n=1}^\infty \tfrac1{m^{n-1}(m-1)} = \tfrac m{(m-1)^2}.$$

Thus $\displaystyle \prod_{n=1}^\infty (1-m^{-n}) > e^{-m/(m-1)^2}.$