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 of the recurrence.
The recurrence is
\[ a_{n+1}=\frac{a_n+6}{a_n+2} \] Let the fixed point be \(L\). Then
\[ L=\frac{L+6}{L+2} \] \[ L(L+2)=L+6 \] \[ L^2+2L=L+6 \] \[ L^2+L-6=0 \] \[ (L-2)(L+3)=0 \] So, the fixed points are
\[ L=2,\,-3 \]

Step 2: Transform the recurrence.
Define
\[ b_n=\frac{a_n-2}{a_n+3} \] Using the recurrence, we get
\[ b_{n+1}=-\frac{1}{5}b_n \] Now,
\[ b_1=\frac{a_1-2}{a_1+3} = \frac{1-2}{1+3} = -\frac14 \] Thus,
\[ b_n=\left(-\frac15\right)^{n-1}\left(-\frac14\right) \]

Step 3: Check convergence.
Since
\[ \left(-\frac15\right)^{n-1}\to 0, \] we get
\[ b_n\to 0 \] Now,
\[ b_n=\frac{a_n-2}{a_n+3} \] So, \(b_n\to 0\) implies
\[ a_n\to 2 \] Therefore, \(\{a_n\}\) is convergent.
Hence, option (C) is true.

Step 4: Check option (A).
For even \(n\), \(b_n>0\).
Also,
\[ |b_{2024}|>|b_{2026}| \] because the magnitude decreases as \(n\) increases.
Since
\[ a_n=\frac{2+3b_n}{1-b_n} \] is an increasing function of \(b_n\), we get
\[ a_{2024}>a_{2026} \] Therefore,
\[ a_{2024}-a_{2026}\geq 0 \] Hence, option (A) is true.

Step 5: Check monotonicity.
The sign of \(b_n\) alternates because of the factor
\[ -\frac15 \] Thus, \(a_n\) alternates around the limit \(2\).
So, even after \(n\geq 9\), the sequence is not monotone.
Hence, option (B) is false.

Step 6: Check option (D).
The largest early value is
\[ a_2=\frac{1+6}{1+2}=\frac73<3 \] Also, the sequence alternates around \(2\) with decreasing magnitude.
Hence, no term exceeds \(3\). Therefore,
\[ a_n\leq 3 \] for all \(n\).
Thus, option (D) is true.

Step 7: Final conclusion.
Hence, the true statements are
\[ \boxed{(A),\,(C)\text{ and }(D)} \]