Question

Let \(\{a_n\}_{n\geq 1}\) be a sequence of real numbers given by

• A. \(a_{2024}-a_{2026}\geq 0\)
• B. \(\{a_n\}_{n\geq 9}\) is a monotone sequence
• C. \(\{a_n\}_{n\geq 1}\) is a convergent sequence
• D. \(a_n\leq 3\) for all \(n\)

Hint:

For recurrence relations of the form \(a_{n+1}=\frac{\alpha a_n+\beta}{\gamma a_n+\delta}\), using a transformation based on fixed points often converts the recurrence into a simple geometric sequence.

Answer 1

The Correct Option is A, C, D

Step 1: Find the fixed points.
The recurrence $a_{n+1}=\frac{a_n+6}{a_n+2}$ has fixed points from $L^2+L-6=0$, namely $L=2$ and $L=-3$.

Step 2: Linearise with a substitution.
Let $b_n=\frac{a_n-2}{a_n+3}$. Then $b_{n+1}=-\frac15 b_n$, a clean geometric chain. With $a_1=1$, $b_1=-\frac14$, so $b_n=-\frac14\left(-\frac15\right)^{n-1}$.

Step 3: Convergence (C).
Since $b_n\to0$, $a_n\to2$, so the sequence converges. (C) holds.

Step 4: Compare $a_{2024}$ and $a_{2026}$ (A).
Both have positive $b$ that shrinks in size, and $a_n$ rises with $b_n$, so $a_{2024}>a_{2026}$, giving $a_{2024}-a_{2026}\ge0$. (A) holds. (B) fails because the sign of $b_n$ alternates, so the sequence is not monotone.

Step 5: Bound the terms (D).
The largest early value is $a_2=\frac73<3$, and terms oscillate toward $2$ with shrinking swings, so $a_n\le3$ always. (D) holds.
\[ \boxed{(A),(C),(D)} \]