4[ Products and factorials

Calculate with five decimal places (i.e., to an accuracy of 0.00001) the product:

$\left(1-\frac{1}{10}\right)\left(1-\frac{1}{10^{2}}\right)\left(1-\frac{1}{10^{3}}\right) \ldots\left(1-\frac{1}{10^{99}}\right)$

#
Let $a=\left(1-10^{-1}\right)\left(1-10^{-2}\right) \ldots\left(1-10^{-5}\right)$ and $b=\left(1-10^{-6}\right) \ldots\left(1-10^{-99}\right)$. Direct calculations show that

$0.89001 + 10^{-6} - x - y$. Therefore, $b > 1 - 10^{-6} - 10^{-7} -$

$\ldots - 10^{-99} > 1 - 2 \cdot 10^{-6}$. Consequently, $ab > 0.890011 \cdot 0.999998 > 0.890009$.

## Answer

$0.89001$.