# Limit of a sequence, with parameter

1. Dec 9, 2012

### Felafel

Could you check my solution please?
1. The problem statement, all variables and given/known data

find out for which values of $\lambda>0$ the sequence $(a_n)$ ,defined by
$a_1 = \frac{1}{2}, \quad\quad a_{n+1} = \frac{1}{2} (\lambda +a_n)^2, \quad n\in \mathbb{N^*}$ converges.
If $(a_n)$ converges, find the limit.

3. The attempt at a solution
If I assume $a_n$ converges, I can write the limit L instead of the elements of the sequence, such that:
$L=\frac{1}{2} \lambda^2 + \frac{1}{2} L^2 + \lambda L$
and
$\lambda^2 + L^2 + 2\lambda L -2L=0$
Solving the equation, I have
$L_1=1-\lambda + \sqrt{1-2\lambda} ; L_2=1-\lambda - \sqrt{1-2\lambda}$
$\lambda$ must therefore be$≤ \frac{1}{2}$
also, $L_1$ must equal $L_2$ due to the theorem of uniqueness of limits.
Thus, the only possible result is $\lambda = \frac{1}{2}$.
And therefore $L=\frac{1}{2}$ too.

2. Dec 9, 2012

### pasmith

I don't think that's right. If the sequence is to converge, then $|a_n - a_{n+1}|$ must tend to 0. But
$$|a_n - a_{n+1}| = \frac12 (\lambda^2 + a_n^2) \geq \frac12 \lambda^2.$$
Clearly the only way for that to tend to zero is if $\lambda = 0$. So if $\lambda = 0$, does the sequence converge?

3. Dec 9, 2012

### Michael Redei

Your condition $\lambda^2+L^2+2\lambda L-2L=0$ is only sufficient (not not necessary) for $L$ being the limit you're looking for. And so the expressions for $L_1$ and $L_2$ are also not necessary, especially not when taken together.

What you need to do is test each expression separately. Is $L_1=1-\lambda+\sqrt{1-2\lambda}$ a possible limit? Then you'd have to have $L_1\leq a_1$ since $(a_n)$ is a falling sequence. If you can show that $L_1\leq a_1$, then that is one limit, which depends on $\lambda$.

Then do the same for $L_2$. Obviously, you can't have two limits, so one of these tests must fail. But the other might not...

4. Dec 11, 2012

### Felafel

okay! so, if $\lambda = \frac{1}{2}, L_1=L_2$ and is the limit, but if $\lambda<\frac{1}{2}$, we have $L_1 > a_1$ which is impossible as the sequence is falling, and the only limit can be $L_2$.
does it work?

5. Dec 11, 2012

### Michael Redei

That's right. Actually, if you look at $L_2$ for $\lambda<\frac12$, you'll get an expression that, if you set $\lambda$ back to $\frac12$ again, should be equal to $\frac12$ itself (if you get the right expression for $L_2$, that is).