## Task B-1.2.

Determine the set of all integers $n$ for which the inequality

$$
\left(1+\frac{1}{3}\right)\left(1+\frac{1}{8}\right)\left(1+\frac{1}{15}\right) \cdots\left(1+\frac{1}{2019^{2}-1}\right)<\frac{2019}{n^{2}}
$$

holds.
## Solution.

$$
\frac{4}{3} \cdot \frac{9}{8} \cdot \frac{16}{15} \cdots \frac{2019^{2}}{(2019-1)(2019+1)}<\frac{2019}{n^{2}}
$$

Each factor on the left side of the equation can be written in the form:

$$
\frac{x^{2}}{x^{2}-1}=\frac{x \cdot x}{(x-1)(x+1)}=\frac{x}{x-1} \cdot \frac{x}{x+1}
$$

Following this, we have:

$$
\begin{aligned}
& \frac{2 \cdot 2}{1 \cdot 3} \cdot \frac{3 \cdot 3}{2 \cdot 4} \cdot \frac{4 \cdot 4}{3 \cdot 5} \cdot \frac{5 \cdot 5}{4 \cdot 6} \cdots \frac{2019 \cdot 2019}{2018 \cdot 2020}<\frac{2019}{n^{2}} \\
& \frac{2}{1} \cdot\left(\frac{2}{3} \cdot \frac{3}{2}\right) \cdot\left(\frac{3}{4} \cdot \frac{4}{3}\right) \cdots\left(\frac{2018}{2019} \cdot \frac{2019}{2018}\right) \cdot \frac{2019}{2020}<\frac{2019}{n^{2}}
\end{aligned}
$$

Notice that each pair of consecutive factors starting from the second to the second-to-last are reciprocals of each other, so their product is 1. After this, we get:

$$
\frac{2 \cdot 2019}{2020}<\frac{2019}{n^{2}}
$$

This inequality holds if and only if $n^{2}<1010$ and $n \neq 0$. This means that:

$$
0<|n|<1010, \quad \text { or } \quad-\sqrt{1010}<n<\sqrt{1010}, n \neq 0
$$

Finally, the integers $n$ that satisfy the condition belong to the set $\{-31,-30, \ldots,-1,1,2, \ldots, 30,31\}$.