Now I want you to solve this math problem: For a given series $a_n$, it satisfies $a_1=c/2, a_{n+1} = (c/2)+(a_n^2/2)$, where c is a constant between 0 and 1. I need to prove that a_n has a limit and find that limit. How do I do that?

Let’s follow Polya’s approach.

Step 1: Understand the problem

We have a sequence defined by: $a_1 = \frac{c}{2}, \quad a_{n+1} = \frac{c}{2} + \frac{a_n^2}{2}$ with ( 0 < c < 1 ).

We need to:

  1. Prove the sequence converges (has a limit as ( $n \to \infty$ )).
  2. Find that limit.

Step 2: Devise a plan

A common strategy for recursively defined sequences:

  1. Check if the sequence is monotonic (increasing or decreasing).
  2. Check if it’s bounded.
  3. If bounded and monotonic, it converges.
  4. Find the limit ( L ) by taking ( $n \to \infty$ ) in the recurrence: ( $L = \frac{c}{2} + \frac{L^2}{2}$ ), then solve for ( L ).

First small step: Can you compute ( $a_2$ ) and compare it to ( $a_1$ ) to guess if the sequence is increasing or decreasing?

I think the sequence is increasing, but I don’t know how to prove it.

Good. Let’s check monotonicity carefully.

We have ( $a_{n+1} - a_n = \frac{c}{2} + \frac{a_n^2}{2} - a_n$ ).

Simplify: $a_{n+1} - a_n = \frac{c + a_n^2 - 2a_n}{2} = \frac{a_n^2 - 2a_n + c}{2}.$

The sign depends on ( $a_n^2 - 2a_n + c$ ).

Question: Can you factor or complete the square for ( $a_n^2 - 2a_n + c$ )?

No. I also checked its delta but it turned out positive. So I can’t prove it is always positive.

Right. The discriminant is ( 4 - 4c > 0 ) since ( c < 1 ), so the quadratic has two real roots.

Let’s find them: $a_n^2 - 2a_n + c = 0 \quad \Rightarrow \quad a_n = 1 \pm \sqrt{1-c}.$

Call ( $r_1 = 1 - \sqrt{1-c}$ ) and ( $r_2 = 1 + \sqrt{1-c}$ ).

Then ( $a_{n+1} - a_n$ ) has the same sign as ( $(a_n - r_1)(a_n - r_2)$ ).

Question: Given ( $a_1 = c/2$ ), where is ( $c/2$ ) relative to ( $r_1$ ) and ( $r_2$ )?
(Hint: ( 0 < c < 1 ), so ( c/2 < 1 ), and ( r_1 < 1 ), ( r_2 > 1 ).)

Oh! I verified that a_1 is smaller than both r_1 and r_2. But what does that imply?

Good. So ( $a_1 < r_1$ ).

Now, if ( $a_n < r_1$ ), then ( $a_{n+1} < r_1$ ) too? Let’s check:

We know ( $a_{n+1} = \frac{c}{2} + \frac{a_n^2}{2} ).
If ( a_n < r_1 ), is ( a_{n+1} < r_1 )?$

Try proving ( $a_{n+1} \le r_1$ ) by using ( $r_1 = 1 - \sqrt{1-c}$ ) and the fact that ( $r_1$ ) satisfies ( $r_1 = \frac{c}{2} + \frac{r_1^2}{2}$ ).

I proved that if a_n is less than r1, than $a_{n+1}$ is also less than r1. So the proof is complete, right? the sequence is monotonically increasing and the limit is r1.

Not quite — you have monotonicity (increasing) and an upper bound ( $r_1$ ), so the sequence converges to some ( $L \le r_1$ ).

But is ( $L = r_1$ ) or ( $L = r_2$ ) or something else?
We find ( L ) by solving ( $L = \frac{c}{2} + \frac{L^2}{2}$ ), which gives ( L = r_1 ) or ( L = r_2 ).

Since ( $a_n < r_1$ ) and increasing, the limit must be ( $r_1$ ).

Final check: Is ( $r_1$ ) actually less than ( $r_2$ )? Yes. So ( $L = r_1$ ).

Do you see any gap in this reasoning?

I don’t see any gap.

Great. Let’s summarize the proof:

  1. Showed ( $a_1 = c/2 < r_1$ ).
  2. Proved by induction: if ( $a_n < r_1$ ), then ( $a_{n+1} < r_1$ ) (bounded above by ( $r_1$ )).
  3. Proved ( $a_{n+1} - a_n > 0$ ) when ( $a_n < r_1$ ) (monotonic increasing).
  4. Bounded increasing sequence → converges to ( L ).
  5. Take limit in recurrence: ( $L = c/2 + L^2/2$ ) → ( $L = 1 \pm \sqrt{1-c}$ ).
  6. Since ( $a_n < r_1$ ) for all ( n ), ( $L \le r_1$ ) → ( $L = r_1 = 1 - \sqrt{1-c}$ ).

You’ve successfully applied Polya’s method: understand, plan, carry out, review. Well done.