[Chain (continuous) fractions]

Solve the equation in positive integers

$$
1-\frac{1}{2+\frac{1}{3+\ldots+\frac{1}{n-1+\frac{1}{n}}}}=\frac{1}{x_{1}+\frac{1}{x_{2}+\ldots \ddots+\frac{1}{x_{n-1}+\frac{1}{x_{n}}}}}
$$
The expression on the left side is greater than $1 / 2$, so $x_{1}<2$, and therefore, $x_{1}=1$. Let

$$
\frac{1}{3+\frac{1}{4+\ldots+\frac{1}{n-1+\frac{1}{n}}}}=a \text { and } \frac{1}{x_{2}+\frac{1}{x_{3}+\cdot \ddots+\frac{1}{x_{n-1}+\frac{1}{x_{n}}}}}=x
$$

Then $1-\frac{1}{2+a}=\frac{1}{1+x}$, which means $1 / x=1+a$. Therefore,

$$
1+\frac{1}{3+\frac{1}{4+\ldots+\frac{1}{n-1+\frac{1}{n}}}}=x_{2}+\frac{1}{x_{3}+\frac{1}{x_{4}+\ldots}+\frac{1}{x_{n-1}+\frac{1}{x_{n}}}}
$$

It is clear that the integer parts of the expressions on the left and right sides are 1 and $x_{2}$, so $x_{2}=1$. After this, we obtain the equality

$$
3+\frac{1}{4+\ldots+\frac{1}{n-1+\frac{1}{n}}}=x_{3}+\frac{1}{x_{4}+\ldots .+\frac{1}{x_{n-1}+\frac{1}{x_{n}}}}
$$

By comparing the integer parts again, we get $x_{3}=3$ and so on.

## Answer

$x_{1}=1, x_{2}=1, x_{3}=3, x_{4}=4, \ldots, x_{n}=n$.