Recursive Sequence Limit

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?

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 $4$.

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