Infimum of recursive sequence $x_1 = 8$ and $x_{n+1} = \frac12 x_n + 2$

Question

The given problem was to find what the given sequence $x_1 = 8$ and $x_{n+1} = \frac12 x_n + 2$ converges to.

First by MCT, I proved that it converges because the sequence is monotonically decreasing (by induction), and that it's bounded below by 4.

The thing now is that I actually want to prove what the sequence converges to, and by MCT I know that's the infimum, but I don't know how to go about actually finding the infimum.

Could someone point me in the right direction?

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Edit: I now know that you don't necessarily need to find the infimum, but I would still like to know how you would approach finding the infimum.

Answer 1

If you know that the sequence converge then $x_{n+1}\to L$ and $x_n\to L$ and backing to the recurrence you get:

$$L=\frac{1}{2}L+2\to 2L=L+4\to L=4$$

Answer 2

Define $\varepsilon_n=x_n-4$ (the subtracted constant being the limit expected from others' arguments) so $\varepsilon_{n+1}=\frac{\varepsilon_n}{2}$. Thus $\varepsilon_n\to 0$, as required.

Answer 3

You can solve the recurrence relation and find the general formula for $x_n$.

The formula for $x_{n+1}=ax_n+b, x_0=c:$

$x_n=\left(c+\frac{b}{a-1}\right)a^n-\frac{b}{a-1}.$

Note that in the given recurrence relation, $x_0=12.$

So: $$x_n=\left(12+\frac{2}{\frac{1}{2}-1}\right)\left(\frac{1}{2}\right)^n-\frac{2}{\frac{1}{2}-1}=8\cdot\left(\frac{1}{2}\right)^n+4.$$

In conclusion, when $n\to\infty, x_n\to4.$