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I'm trying to show that the limit of the following recursive sequence is 4.

$a_{n+1} = \frac{1}{2}a_{n} + 2$ and $a_{1} = \frac{1}{2}$.

Could someone give me a hint as to how to start this problem? I've been stuck on this for a while.

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I belive your formula should be $a_{n+1} = \frac{1}{2}a_n + 2$, otherwise there the problem isn't well-posed (and my "fixed" sequence does indeed go to 4) – Tyler Mar 5 '12 at 5:40
(More for the benefit of other readers than the OP) If you can't be bothered applying induction twice to show it's bounded and monotone, the Banach contraction mapping theorem destroys this. – Ragib Zaman Mar 5 '12 at 10:43
up vote 7 down vote accepted

So first, you need to show it converges.

Induction will tell you every term is bounded by $4$. Induction can also tell you that the sequence is monotone increasing. Thus the sequence converges to some number.

Then note that $a_n/2 + 2 = \frac{a_n + 4}{2}$, so that you are constantly taking the arithmetic average of your terms with $4$.

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Very helpful. I would vote up but I don't have the reputation yet. – user26139 Mar 5 '12 at 5:46

Another way to prove the limit is $4$...

You know by mixedmath's answer that the sequence is bounded above and increasing, so there is a (finite) limit, say, $L$. Take the limit on both sides:

$\lim \limits_{n \to \infty} a_{n+1} = \lim \limits_{n \to \infty} (\frac{1}{2}a_n + 2)$


$L = \frac{1}{2}L + 2$

which means

$L = 4$.


This method can generalize to find limits of other recursively defined functions, for example, consider the following equation:

$a_{n+1} = \sqrt{2 + a_n}$ and $a_0 = \sqrt{2}$

Can you prove the limit exists, and using the method above find the value?

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Clear and concise, thanks! – user26139 Mar 5 '12 at 5:46
If I use the method you suggested to solve my problem, I would need to solve $L^{2} - L - 2 = 0$. Using the quadratic formula I get that $L = -1$ or $L = 2$. But since the sequence is bounded below by $\sqrt{2}$, the limit $L = 2$. – user26139 Mar 5 '12 at 5:54

$4 - a_{n+1} = 2 - \frac 1 2 a_n$, so

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

$4 - a_1 = 3 \frac 1 2$

So we can have a function $f_n$ on the natural numbers such that:

$f_n = 4 - a_n$

By the recursions above, we show that $f_n = 7 \left( \frac 1 2 \right)^n$. Now you must show that this function approaches $0$ as $n$ approaches infinity.

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Very interesting apporach! – Pedro Tamaroff Mar 6 '12 at 2:09
@Andrew Salmon: Very nice! – Student Mar 6 '12 at 2:44

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