17.3. (New York, 74). Let

$$
a_{n}=\frac{1 \cdot 3 \cdot 5 \ldots(2 n-1)}{2 \cdot 4 \cdot 6 \ldots 2 n}, n \in \mathrm{N}
$$

Find $\lim a_{n}$.

$$
n \rightarrow \infty
$$
17.3. Since

$$
\begin{aligned}
a_{n}^{2}=\frac{1^{2} \cdot 3^{2} \cdot \ldots \cdot(2 n-1)^{2}}{2^{2} \cdot 4^{2} \cdot \ldots \cdot(2 n)^{2}} & = \\
& =\frac{1 \cdot 3}{2^{2}} \cdot \frac{3 \cdot 5}{4^{2}} \cdots \frac{(2 n-1)(2 n+1)}{(2 n)^{2}} \cdot \frac{1}{2 n+1}<\frac{1}{2 n+1}
\end{aligned}
$$

for any $n \in \mathbf{N}$, then

$$
0<a_{n}<1 / \sqrt{2 n+1} \quad \text { and } \quad \lim _{n \rightarrow \infty} a_{n}=0
$$