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?
 A: 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
$$
\lim_{n\to\infty} a_n=4.
$$
A: 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 $0.5+3.5=4$.
A: Use contractive condition to prove it is Cauchy  sequence which will imply it converges:
Use: $|a(n+2) - a(n+1)| < k|a(n+1) - a(n)|$
If $0 < k < 1$ then it's a Cauchy  sequence.
On solving $k$ comes out to to be a half (so, $1/2$).
Hence sequence converges.
To find limit: say limit is $x$.
As $n$ tends to infinity so $a(n+1)$ is almost equal to $a(n)$, so putting $x$ in place of $a(n)$ and $a(n+1)$ that is:
$$x = x/2 + 2$$
so limit is $x=4$.
A: 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 $y=0$
A: The solution to this recurrence is 
$$a_n=4-\frac{7}{2^n},$$ 
so $a_n$ converges to $4$.
