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Limit of a sequence, with parameter

  • Thread starter Felafel
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  • #1
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Could you check my solution please?

Homework Statement



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.

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.
 

Answers and Replies

  • #2
pasmith
Homework Helper
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Could you check my solution please?

Homework Statement



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.

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.
I don't think that's right. If the sequence is to converge, then [itex]|a_n - a_{n+1}|[/itex] must tend to 0. But
[tex]
|a_n - a_{n+1}| = \frac12 (\lambda^2 + a_n^2) \geq \frac12 \lambda^2.
[/tex]
Clearly the only way for that to tend to zero is if [itex]\lambda = 0[/itex]. So if [itex]\lambda = 0[/itex], does the sequence converge?
 
  • #3
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
171
0
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...
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
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?
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).
 

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