# Proving convergence of a sequence

Let the following recursively defined sequence:

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

$a_1=\dfrac{1}{2}$.

Prove that $a_n$ converges to 4 by subtracting 4 from both sides.

When I do that, I get:
$2(\frac{1}{2} a_{n+1} -2)=(\frac{1}{2} a_n -2)$, so $y=2y$,
which is true only for $0$. But I’m not sure how to formally use this in a definition of convergence?

#### Solutions Collecting From Web of "Proving convergence of a sequence"

Set $b_n=a_n-4$. Then $b_1=-7/2$ and
$$b_{n+1}=a_{n+1}-4=\frac{1}{2}a_n+2-4=\frac{1}{2}(a_n-4)=\frac{1}{2}b_n.$$
Thus
$$b_n=\frac{b_{n-1}}{2}=\frac{b_{n-2}}{2^2}=\cdots=\frac{b_{1}}{2^{n-1}}=-\frac{7}{2^n},$$
and finally
$$a_n=4-\frac{7}{2^n}.$$
Hence
$$\lim_{n\to\infty} a_n=4.$$

Just a generalization of the approach for you. Note that
$$a_n=\frac{1}{2}a_{n-1}+2$$
So if you subtract this expression from the $a_{n+1}$ you have above, you get rid of the constant. Also denote $\Delta a_{n+1}=a_{n+1}-a_{n}$ and you get:

$$\Delta a_{n+1}=\frac{1}{2} \Delta a_{n}=\frac{1}{2^2} \Delta a_{n-1}=\ldots =\frac{1}{2^{n-1}} \Delta a_{2}$$

If you sum over $n$ the LHS you get a telescoping sum: $\sum_{n=1}^{N}a_{n+1}=a_{N+1}-a_1$. Since $a_2=2 \frac{1}{4}$ and $a_1 = 0.5$ you get (using geometric sum $\sum_{n=1}^{N} \frac{1}{2^{n-1}}=2(1-(\frac{1}{2})^{N+1})$

$$a_{N+1}=0.5+(2.25-0.5) \cdot 2 \cdot \Bigg(1-(\frac{1}{2})^{N+1} \bigg)$$
and if you take the limit as $N \to \infty$ you get $0.5+3.5=4$.

You have $2(\frac{1}{2}a_{n+1}-2)=\frac{1}{2}a_{n}-2$. So, you have a sequence $y_n$ where $y_{n+1}=\frac{1}{2}y_n$. This is a bounded decreasing sequence, and hence has a limit, say $y$. Then $y$ satisfies $y=\frac{1}{2}y$ so $y=0$

The solution to this recurrence is
$$a_n=4-\frac{7}{2^n},$$
so $a_n$ converges to $4$.