Solving an equation (2)

**NonCommAlg** · December 12th 2009, 05:00 PM

Find all real numbers $x \geq 0$ such that $\prod_{n=2}^{\infty} \left(1 - \frac{1}{n(\ln n)^x} \right) = 0.$

**chisigma** · December 13th 2009, 12:42 PM

The 'infinite product' ...

$\prod _{n} (1-a_{n})$ (1)

... diverges if the corresponding 'infinite sum'...

$\sum _{n} a_{n}$ (2)

... diverges. In this case the 'infinite sum' is...

$\sum _{n=2}^{\infty} \frac{1}{n\cdot \ln^{x} n}$ (3)

... and we can use the 'integral test' to verify its convergence. If $x=1$ we have...

$\int\frac{1}{t\cdot ln t} \cdot dt = \ln \ln |t| + c$ (4)

... and the integral from $2$ to $\infty$ diverges. If $x>1$ we have...

$\int\frac{1}{t\cdot ln^{x} t} \cdot dt = \frac{\ln^{1-x} t}{1-x} + c$ (5)

... and the integral from $2$ to $\infty$ converges. The conclusion is that the 'infinite product'...

$\prod _{n=2}^{\infty} (1-\frac{1}{n\cdot \ln^{x} n})$ (6)

... converges for $x>1$ and diverges for $x\le 1$ ...

Merry Christmas from Italy

**Wilmer** · December 13th 2009, 01:10 PM

Originally Posted by chisigma

My new avatar will remain until the sentence of the European Court that removes the crucifix from the italian schools will be revised...

I thought there was a rule here preventing "religious views". No?

**redsoxfan325** · December 13th 2009, 01:43 PM

Originally Posted by chisigma

The 'infinite product' ...

$\prod _{n} (1-a_{n})$ (1)

... diverges if the corresponding 'infinite sum'...

$\sum _{n} a_{n}$ (2)

... diverges. In this case the 'infinite sum' is...

$\sum _{n=2}^{\infty} \frac{1}{n\cdot \ln^{x} n}$ (3)

... and we can use the 'integral test' to verify its convergence. If $x=1$ we have...

$\int\frac{1}{t\cdot ln t} \cdot dt = \ln \ln |t| + c$ (4)

... and the integral from $2$ to $\infty$ diverges. If $x>1$ we have...

$\int\frac{1}{t\cdot ln^{x} t} \cdot dt = \frac{\ln^{1-x} t}{1-x} + c$ (5)

... and the integral from $2$ to $\infty$ converges. The conclusion is that the 'infinite product'...

$\prod _{n=2}^{\infty} (1-\frac{1}{n\cdot \ln^{x} n})$ (6)

... converges for $x>1$ and diverges for $x\le 1$

I think I'm not understanding something. How do you know it converges to 0?

**chisigma** · December 13th 2009, 07:50 PM

Originally Posted by redsoxfan325

I think I'm not understanding something. How do you know it converges to 0?...

A precise enough definition of 'infinite product' is given here...

Infinite product - Wikipedia, the free encyclopedia

If You have an infinite product in the form...

$\prod_{n=1}^{\infty} a_{n} = \lim_{n \rightarrow \infty} \prod_{k=1}^{n} a_{k}$ (1)

... it is said to be convergent if the limit exists and it is not zero. In all other cases it is said to be divergent. If $\forall n , a_{n}>0$ we can write...

$\ln \prod_{n=1}^{\infty} a_{n} = \sum_{n=1}^{\infty} \ln a_{n}$ (2)

... so that the convergence of the infinite product is equivalent of the convergence of the 'infinite sum' of logarithms. If a product of infinite positive terms diverges the limit can be $0$ or $+ \infty$ . An useful criterion of convergence of an infinite product is allowable when the general term can be written as $a_{n}= 1 + p_{n}$ . In this case the (1) converges if converges the 'infinite sum'...

$\sum_{n=1}^{\infty} p_{n}$ (3)

Merry Christmas from Italy

$\chi$ $\sigma$

**redsoxfan325** · December 13th 2009, 07:54 PM

Oh, okay. I didn't realize that a product of zero was considered divergent. (You can imagine my confusion when I read your post not knowing that!)

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