<s>[INST] 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?
 [/INST] 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 $0.5+3.5=4$.
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