How can I systematically find a solution to this problem? While doing some self-study, a friend posed this problem to me:

Let $a$ be a sequence, defined as following:
$$a_0 = 0,\quad a_1= 1, \qquad a_{n+2}=\frac{a_n+a_{n+1}}2$$
Figure out, whether $a$ converges, and if yes to which value. Find a closed form for $a_n$. If you want to, you may also have a look at the common case with $a_0= \alpha$ and $a_1 = \beta$.

It's easy to see, that this converges to $\frac2 3$, and the other parts are also easy to solve with quessing a result and afterwards proving it correct, but I dislike this approach, as it is not productive if the solution isn't obvious.
Is there a systematically approach for such problems? I would be really happy if somebody can explain me, how to systematically solve problems like this.
 A: There are several.  My favorite, and one of the most generalizable, is the use of Generating functions.  These offer a systematic way of solving many recurrence relations of any order.  (Yours is a 2nd order equation)
An excellent place to start learning about them is this online book.
Hope that helps.
A: Less often effective than generating functions, but in my opinion less work, is to guess that the solution to a linear homogeneous recurrence is of the form ar^n.  The a is like a constant of integration  Plug it in and in your case you get $ar^2=(a+ar)/2$ or $2r^2-r-1=0$.  There are two roots, $r_1$ and $r_2$ so the solution is $ar_1^n+br_2^n$  Now use your initial values to find $a$ and $b$.  If you have a repeated root you will have a term in $anr^n$ as well.
A: If the recurrence relation is linear, homogeneous and has constant coefficients, here is a systematic way to solve it. Similar method will work if you have linear, inhomogeneous equation with constant coefficients. (For instance if the inhomogeneity is a constant, the inhomogeneity can be removed if possible by defining $b_n = a_n + c$ and figuring out $c$ such that the equation in $b_n$ is homogeneous. If it is not possible, which happens when the sum of the coefficients of $a_n$ add up to zero, then there is a linear growth term added to $a_n$)
First obtain the characteristic equation. To do this, assume $a_n = m^n$.
Plug it in to get a quadratic in $m$ in this case. Solve for $m$. (If you have $a_n$ depending on $a_{n-1},a_{n-2},\ldots,a_{n-p}$, you will get a $p^{th}$ order polynomial.)
Get the two roots say $m_1, m_2$. (In general, the $p$ roots).
Now the general solution is given by the linear combination of the roots namely $a_n = c_1 m_1^n + c_2 m_2^n$. (In general, $a_n = \displaystyle \sum_{k=1}^p c_k m_k^n$)
Solve for $c_1$ and $c_2$ (In general, $c_1,c_2,\ldots,c_p$) using the initial conditions
For your recurrence, the corresponding equation becomes, $2m^2-m-1 = 0 \Rightarrow (2m+1)(m-1) = 0 \Rightarrow m = -\frac{1}{2},1$
Hence, $a_n = c (-\frac{1}{2})^n + d$
If $a_0 =0, a_1 =1$, we get $c + d = 0$ and $d - \frac{c}{2} = 1$
$d = \frac{2}{3}$ and $c = -\frac{2}{3}$
Hence, $a_n = \frac{2}{3} - \frac{2}{3} (-\frac{1}{2})^{n}$
Hence, $\displaystyle \lim_{n \rightarrow \infty} a_n = \frac{2}{3}$
If $a_0 =\alpha, a_1 =\beta$, we get $c + d = \alpha$ and $d - \frac{c}{2} = \beta$
$c = \frac{2}{3}(\alpha-\beta)$, $d = \frac{\alpha + 2 \beta}{3}$
Hence, $a_n = \frac{\alpha + 2 \beta}{3} + \frac{2}{3}(\alpha-\beta) (\frac{-1}{2})^n$
Hence, $\displaystyle \lim_{n \rightarrow \infty} a_n = \frac{\alpha + 2 \beta}{3}$
This methodology is analogous to plugging in $y=e^{mx}$ when you want to solve a linear, homogeneous ODE with constant coefficients. You get the constants for the ODE using the boundary conditions. Here you have a difference equation instead of a differential equation.
A: I realize you're asking for a systematic answer to the problem, and others have supplied such, but there is a neat correspondence with the binary representation of 1/3. 
Every iteration of the recurrence, you're dividing everything by two each time, you're going up and down each time by a negative power of 2, $1 - 1/2 + 1/4 - 1/8 + ...$ or $1/2 + 1/8 + 1/32 + ...$ essentially creating a binary encoding .1010101... which, as a geometric series comes out to $${1\over 2}\cdot{1\over 1- {1\over 4}} = {2\over 3}.$$
