Question

Let \( \{a_n\}_{n \geq 1} \) be a sequence of real numbers such that \( a_1 = 3 \) and, for \( n \geq 1 \), \[ a_{n+1} = \frac{a_n^2 - 2a_n + 4}{2}. \] Then which of the following statement(s) is (are) true?

• A. \( \{a_n\}_{n \geq 1} \) is a monotone sequence
• B. \( \{a_n\}_{n \geq 1} \) is a bounded sequence
• C. \( \{a_n\}_{n \geq 1} \) does not have finite limit, as \(n \to \infty\)
• D. \( \lim_{n \to \infty} a_n = 2 \)

Hint:

For recurrence relations, check for fixed points (where the sequence stabilizes) and examine the behavior of the terms to determine convergence.

Answer 1

The Correct Option is A, C

Step 1: Compute the first few terms

$$a_1 = 3$$ $$a_2 = \frac{3^2 - 2(3) + 4}{2} = \frac{9 - 6 + 4}{2} = \frac{7}{2} = 3.5$$ $$a_3 = \frac{(3.5)^2 - 2(3.5) + 4}{2} = \frac{12.25 - 7 + 4}{2} = \frac{9.25}{2} = 4.625$$ $$a_4 = \frac{(4.625)^2 - 2(4.625) + 4}{2} = \frac{21.390625 - 9.25 + 4}{2} = \frac{16.140625}{2} \approx 8.07$$

The sequence appears to be increasing: $3 < 3.5 < 4.625 < 8.07 < ...$

Step 2: Find fixed points

If the sequence converges to a limit $L$, then: $$L = \frac{L^2 - 2L + 4}{2}$$ $$2L = L^2 - 2L + 4$$ $$L^2 - 4L + 4 = 0$$ $$(L - 2)^2 = 0$$ $$L = 2$$

The only fixed point is $L = 2$.

Step 3: Analyze monotonicity

We need to check if $a_{n+1} > a_n$ or $a_{n+1} < a_n$.

$$a_{n+1} - a_n = \frac{a_n^2 - 2a_n + 4}{2} - a_n = \frac{a_n^2 - 2a_n + 4 - 2a_n}{2} = \frac{a_n^2 - 4a_n + 4}{2} = \frac{(a_n - 2)^2}{2}$$

Since $(a_n - 2)^2 \geq 0$ for all $a_n$, we have $a_{n+1} - a_n \geq 0$.

Moreover, $a_{n+1} = a_n$ only if $a_n = 2$.

Since $a_1 = 3 \neq 2$, we have $a_2 > a_1$, and by induction, $a_n > 2$ for all $n \geq 1$.

Therefore, $a_{n+1} > a_n$ for all $n \geq 1$, making the sequence strictly increasing.

Option (A) is TRUE 

Step 4: Check if the sequence is bounded

Since $a_n > 2$ for all $n$ and $a_{n+1} - a_n = \frac{(a_n - 2)^2}{2} > 0$, the sequence is strictly increasing.

Let's check if it's bounded above. For large $a_n$: $$a_{n+1} = \frac{a_n^2 - 2a_n + 4}{2} \approx \frac{a_n^2}{2}$$

This suggests exponential growth. Let's verify: if $a_n = c$ for large $c$: $$a_{n+1} \approx \frac{c^2}{2}$$

This grows without bound, so the sequence is unbounded.

Option (B) is FALSE 

Step 5: Determine if the sequence has a finite limit

Since the sequence is strictly increasing and unbounded above, it diverges to infinity: $$\lim_{n \to \infty} a_n = \infty$$

The sequence does not have a finite limit.

Option (C) is TRUE 

Option (D) is FALSE  (since the limit is $\infty$, not 2)

Answer: (A) and (C)