Finding the limit of the sequence $T_1=0,T_2=1, T_n=\frac{T_{n-1}+T_{n-2}}{2}$ Given that $T_1=0$, $T_2=1$ and $T_n=\frac{T_{n-1}+T_{n-2}}{2}$, show that the sequence converges to $\frac{2}{3}$.

I am quite a fan of the following solution: We fist show that the sequence converges for arbitrary two starting values and then derive the limit indirectly from its dependence on the starting values.
Let $a,b \in \mathbb{R}$ and define the sequence $(c_n)_{n \in \mathbb{N}}$ by $c_0 = a$, $c_1 = b$ and $c_{n+2} = (c_n + c_{n+1})/2$ for all $n \in \mathbb{N}$. Then for all $n \geq 1$ we have
$$
 | c_{n+1} - c_{n} |
 = \left| \frac{c_n + c_{n-1}}{2} - c_n \right|
 = \frac{|c_n-c_{n-1}|}{2}
$$
and thus
$$
 | c_{n+1} - c_{n} |
 = \frac{|c_n-c_{n-1}|}{2}
 = \frac{|c_{n-1}-c_{n-2}|}{4}
 = \dotsb
 = \frac{|c_1-c_0|}{2^n}
 = \frac{|b-a|}{2^n}.
$$
It follows that for all $m \geq n \geq 1$ we have
$$
 |c_m - c_n|
 \leq \sum_{k=n}^{m-1} |c_{k+1}-c_k|
 = \sum_{k=n}^{m-1} \frac{|b-a|}{2^k}
 \leq \sum_{k=n}^\infty \frac{|b-a|}{2^k}
 = \frac{|b-a|}{2^{n-1}}.
$$
Thus the sequence $(c_n)_{n \in \mathbb{N}}$ is a Cauchy sequence and thus converges.
We can now calculate the limit in a rather nice way: For all $a,b \in \mathbb{R}$ let $\Phi(a,b) \in \mathbb{R}$ denote the limit of the above sequence with start values $a$ and $b$. We want to show that $\Phi(0,1) = \frac{2}{3}$.
For this notice that from the properties of convergent series and the special form of our sequence it directly follows that for all $a,b,c \in \mathbb{R}$ we have
$$
 \Phi(a+c,b+c) = \Phi(a,b) +c, \quad
 \Phi(ca,cb)= c\, \Phi(a,b), \quad
 \Phi(a,b) = \Phi\left(b, \frac{a+b}{2}\right)
$$
From this we can now follow that
\begin{align*}
 \Phi(0,1)
 = \frac{1}{2} \Phi(0,2)
 = \frac{1}{2} \Phi(2,1)
 = \frac{1}{2} \Phi(0,-1) + 1
 = -\frac{1}{2} \Phi(0,1) + 1.
\end{align*}
Comparing the left and right hand side of the above equality yields $\Phi(0,1) = \frac{2}{3}$.
From this we find even more generally that for all $a,b \in \mathbb{R}$ we have
$$
 \Phi(a,b)
 = \Phi(0,b-a) + a
 = (b-a) \Phi(0,1) + a
 = \frac{2}{3}(b-a) + a
 = \frac{2b+a}{3}.
$$
