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