# Thread: Solving an equation (2)

1. ## Solving an equation (2)

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

2. The 'infinite product' ...

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

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

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

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

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

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

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

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

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

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

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

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

Merry Christmas from Italy

3. 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?

4. Originally Posted by chisigma
The 'infinite product' ...

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

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

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

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

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

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

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

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

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

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

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

... converges for $\displaystyle x>1$ and diverges for $\displaystyle x\le 1$
I think I'm not understanding something. How do you know it converges to 0?

5. 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...

$\displaystyle \prod_{n=1}^{\infty} a_{n} = \lim_{n \rightarrow \infty} \prod_{k=1}^{n} a_{k}$ (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 $\displaystyle \forall n , a_{n}>0$ we can write...

$\displaystyle \ln \prod_{n=1}^{\infty} a_{n} = \sum_{n=1}^{\infty} \ln a_{n}$ (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 $\displaystyle 0$ or $\displaystyle + \infty$. An useful criterion of convergence of an infinite product is allowable when the general term can be written as $\displaystyle a_{n}= 1 + p_{n}$. In this case the (1) converges if converges the 'infinite sum'...

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

Merry Christmas from Italy

$\displaystyle \chi$ $\displaystyle \sigma$

6. 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!)