# Problem 5.

Given numbers $x_{1}, x_{2}, x_{3}, \ldots, x_{2018}$. It is known that $x_{1}=1 / 2$ and

$$
x_{n}=\frac{x_{n-1}}{2 n x_{n-1}+1} \text { for } n=2, \ldots, 2018
$$

Find the sum $x_{1}+x_{2}+\cdots+x_{2018}$.

#
# Solution.

Let's introduce the sequence $y_{n}=1 / x_{n}$. For this sequence, the recurrence relation can be rewritten as

$$
y_{n}=y_{n-1}+2 n
$$

Thus,

$$
y_{n}=2+4+\ldots+2 n=n(n+1)
$$

from which we get

$$
x_{n}=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}
$$

Summing such terms for all $\mathrm{n}$ from 1 to 2018, we obtain that the sum is

$$
1-\frac{1}{2019}=\frac{2018}{2019}
$$

Answer: $\frac{2018}{2019}$.