Recursive Sequence Limit 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.
 A: 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 $4$.
A: $4 - a_{n+1} = 2 - \frac 1 2 a_n$, 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.
A: 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:
$\lim \limits_{n \to \infty} a_{n+1} = \lim \limits_{n \to \infty} (\frac{1}{2}a_n + 2)$
So,
$L = \frac{1}{2}L + 2$
which means
$L = 4$.
EDIT:
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?
A: Take $f(x) = \frac{1}{2}x+2$, then $f: [1/2, 4] \to [1/2, 4]$ as easy calculations show.
Moreover $|f(x)-f(y)| = \frac{1}{2}|x-y|$ so by Banach's fixed point theorem $x_n = f^{n-1}(1/2)\to L$ where $f(L) = L$ that is $L = 4$.
