[ Sequences (other). ] [ Identical transformations ]

## Find the largest term of the sequence $x_{n}=\frac{n-1}{n^{2}+1}$.
Compare two adjacent members of the sequence.

## Solution

Let's compare two adjacent members of the sequence. For this, we write the difference \( x_{n+1} - x_{n} = \frac{n}{(n+1)^2 + 1} - \frac{n-1}{n^2 + 1} \). After identical transformations, we get: \( x_{n+1} - x_{n} = -\frac{n^2 - n - 2}{((n+1)^2 + 1)(n^2 + 1)} = -\frac{(n-2)(n+1)}{((n+1)^2 + 1)(n^2 + 1)} \). Thus, the difference \( x_{n+1} - x_{n} \) is greater than 0 when \( n < 2 \), equal to 0 when \( n = 2 \), and less than 0 when \( n \geq 3 \). Therefore, \( x_{1} < x_{2} = x_{3} > x_{4} > x_{5} > \ldots \), and \( x_{2} = x_{3} = 0.2 \) are the largest members of this sequence.