Compute upper limit and lower limit Consider the sequence, $0<a_{1}<a_{2}$ and
\begin{eqnarray*}
 a_{n}= \frac{a_{n-1}+a_{n-2}}{2}
\end{eqnarray*}
for $n\geq 3$. Show that:
\begin{eqnarray*}
\overline{\lim a_{n}}=\underline{\lim a_{n}}
\end{eqnarray*}

For this, the bounds are $a_{1}$ and $a_{2}$ so I can describe the sequence:
\begin{align*}
\overline{a_{k}} &= \begin{cases}
             a_{k} & \text{if } k \text{ is even} \\
             a_{k+1} & \text{if } k \text{ is odd}
             \end{cases} \\
\underline{a_{k}} &= \begin{cases}
             a_{k} & \text{if } k \text{ is odd}  \\
             a_{k+1} & \text{if } k \text{ is even}
             \end{cases}
\end{align*}
So $\{\overline{a_{k}}\}=\{a_{2},a_{4},a_{6},...\}$ and $\{\underline{a_{k}}\}=\{a_{1},a_{3},a_{5},...\}$, the sequence $\{\overline{a_{k}}\}$ is decreasing and  the sequence $\{\underline{a_{k}}\}$ is increasing, so, by Weirstrass's theorem are convergent but when I try to compute $\overline{\lim a_{n}}$ or $\underline{\lim a_{n}}$ I try to do this, Since:
\begin{eqnarray*}
a_{2k}=\frac{a_{2k-1}+a_{2k-2}}{2}
\end{eqnarray*}
for $n\geq 3$ and $\lim a_{2k}= x = \lim a_{2k-1}= \lim a_{2k-2}$, but this don't say a great information, Can you give some hint or advice to start? Thank you
 A: At each step, you take the center of the two previous terms. That means that $a_3$ is the center of $[a_1,a_2]$, samely $a_4$ is the center of $[a_2,a_3]$, etc... This suggests that $a_{n+1}-a_n$ behaves like $\frac{1}{2^n}$, indeed
$$ a_{n+2}-a_{n+1}=\frac{a_{n+1}+a_n}{2}-a_{n+1}=-\frac{1}{2}(a_{n+1}-a_n) $$
Therefore $a_{n+1}-a_n=\lambda\left(-\frac{1}{2}\right)^n$ where $\lambda$ is a constant. Summing this gives that $a_n=\alpha+\beta\left(-\frac{1}{2}\right)^n$ where $\alpha,\beta$ are constants. Therefore $(a_n)$ converges and thus $\liminf\limits_{n\rightarrow +\infty} a_n=\limsup\limits_{n\rightarrow +\infty} a_n=\lim\limits_{n\rightarrow +\infty} a_n$.
Note : You can directly show that $a_n=\alpha+\beta\left(-\frac{1}{2}\right)^n$ by searching the roots of $X^2-\frac{1}{2}X-\frac{1}{2}$, which are $1$ and $-\frac{1}{2}$.
A: Just rewriting some of what you wrote, plus some of my comment, to make it all easier to read. So, we have:
$0<a_{1}<a_{2}$ and
$\displaystyle a_{n}= \frac{a_{n-1}+a_{n-2}}2,$
for $n\geq 3$.
That is, $\displaystyle a_3= \frac{a_2+a_1}2$ so
$0<a_1<a_3<a_2.$
Similarly, $\displaystyle a_4= \frac{a_3+a_2}2$ so
$0<a_1<a_3<a_4<a_2,$  etc.
As you observed, the subsequence with even indices is decreasing (and clearly bounded), hence converging. Let $L=\lim\limits_{k\to\infty} a_{2k}$.
Then clearly $a_{2k-1}\le L$ for all $k$, hence if we let $l=\lim\limits_{k\to\infty} a_{2k-1}$ then we have $l\le L$. (The limit $l$ exists since the subsequence with odd indices is increasing and bounded.)
Now it is enough to show that $l=L.$
As you wrote $\displaystyle a_{2k}=\frac{a_{2k-1}+a_{2k-2}}{2}.$
So, if we let $k\to\infty$ we obtain:
$\displaystyle L=\frac{l+L}{2}.$
But then, $\displaystyle \frac L2=L-\frac L2=\frac l2,$ hence $L=l.$
It follows that $\lim\limits_{k\to\infty} a_k=L=l.$
