$0Let $a_n$ be a sequence of positive real numbers such that
$$a_n<\frac{a_{n-1}+a_{n-2}}{2}$$
Show that $a_n$ converges.
 A: Look at the sequence given by $b_n := \max (a_{n-1},a_{n-2})$. Now, $a_n ≤ b_n$ for all $n ∈ ℕ$, and $(b_n)$ is monotonically nonincreasing, because
$$b_{n+1} = \max (a_n,a_{n-1}) ≤ \max(b_n, a_{n-1}) = b_n.$$
Since $(b_n)$ is nonnegative nonincreasing, it has to converge and therefore so does $(a_n)$, since we can sandwich.
This argument is incomplete, as pointed out in the comments: What is the other part of the sandwich? However, Did did complete the argument in the comments.
A: the following line of reasoning may be a step towards a solution:
with a slight change of notation, define $s_0=a_0$ and for $n \ge 1$ let $s_n=a_n+\frac12 a_{n-1}$. the question is now: if $\{s_k\}$ is a convergent, decreasing sequence of non-negative reals, does the sequence $\{a_k\}$ converge? 
describe the sequences up to their $n$-th terms as $n+1$-vectors $S_n$ and $A_n$. then we can define a sequence of square matrices $Q_n$, such that:
$$ S_n = Q_n A_n
$$
here, obviously, $Q_0$ is just the scalar value $1$, and each subsequent $Q_n$ is an $n \times n$ matrix whose diagonal elements are all $1$, and whose only other non-zero entries are $q_{i+1,i}=\frac12$, for $i=0,\dots,n-1$.
the matrices $Q_n$ are nonsingular, with inverses $Q_n^{-1}$, whose elements are given by (NB subscripts range from $0$ to $n$)
$$ q_{ij}^{-1}= \left( \frac{-1}{2} \right)^{j-i}
$$ for $i \ge j$ and zero otherwise. this gives:
$$
a_n = s_n - \frac12 s_{n-1} + \frac14 s_{n-2} -\dots + \left( \frac{-1}{2} \right)^n s_0
$$
if the limit of the sequence $s_n$ is $\lambda$ then:
$$
a_n - \frac23 \lambda = \sum_{k=0}^n (s_{n-k}-\lambda) \left(\frac{-1}2\right)^k \\
=  \sum_{k=0}^{n-N} (s_{n-k}-\lambda) \left(\frac{-1}2\right)^k +\left(\frac{-1}2\right)^{n-N}\sum_{k=1}^{N}(s_{n-N+k}-\lambda)\left(\frac{-1}2\right)^k
$$
A: The same holds for a nonlinear average as the upper bound on $a_n$.
Suppose that $a_n \leq f(a_{n-1},a_{n-2}, \dots, a_{n-k})$ for a generalized mean $f$, which for this answer means:


*

*$f(x_1,\dots,x_k)$ is increasing in all its variables, and continuous 

*$f(a,a,a,\dots,a)=a$ for all $a$

*$f$ has finite velocity, defined here to mean that $\partial f / \partial{x_i} \geq c > 0$ for a constant $c$ and all $i=1,2,...,k$ 


(It is not assumed that $f$ is symmetric in its variables, or differentiable. The finite velocity condition is stated with derivatives only to simplify notation.)
Then: if the sequence $a_n$ is bounded below, it converges.
Proof.
1) $m_n = \max(a_{n},a_{n+1},\dots a_{n+k-1})$ satisfies $m_n \geq m_{n+k}$
2) sequence $m_{kn}+r$ has a limit $M_r$ for any integer $r$ and $n \to \infty$, because it is decreasing and bounded below
3) there are a finite number of values of $M_r$, which is a periodic function of $r$ with period $k$, but they are all equal, because the largest $M_r$, call it $\mathbb{M}$, must (by finite velocity) come from maxima of consecutive $k$-tuples $(a_i, a_{i+1}, \dots, a_{i+k-1})$ that are infinitesimally close to $\mathbb{M}$; then the next $k$ values ($a_{i+k}$ to $a_{i + 2k-1}$) are also (by continuity of $f$) close to $\mathbb{M}$, and a block of $2k$ values infinitesimally close to $\mathbb{M}$ means that all $M_r$ are also at least that close to $\mathbb{M}$ (by generalized mean, $f$ is between the min and max of its arguments), hence equal.
4) For large $n$ the $a_n$ converge to $\mathbb{M}$, since this is the case on the progressions $n = r \mod k$ (or repeat the same argument as in (3), taking blocks of $k$ terms whose maxima are close to $\mathbb{M}$).
The essential point is that finite velocity rules out oscillations between different limiting values on the subsequences, as would happen with $f(x,y)=x$.
A: does the following approach lead anywhere? we have $$ a_{n+2}=\frac{a_{n+1}+a_n}{2} - \epsilon_n$$
where $0 \lt \epsilon_n \lt \frac{a_{n+1}+a_n}{2}$.  so by adding,
$$  a_{n+2}+a_{n+1} - a_1-a_0 = \frac12 (a_{n+1} - a_0) - \sum_{k=0}^n \epsilon_k
$$ or
$$ a_{n+2}+\frac12 a_{n+1} = a_1+\frac12 a_0 - \sum_{k=0}^n \epsilon_k
$$ defining:
$$s_{n+1} = a_{n+2} +\frac12 a_{n+1}$$
this becomes
$$s_{n+1} = s_0 - \sum_{k=0}^n \epsilon_k
$$
which guarantees convergence of $s_n$
A: Set $$b_n=a_n+\frac{1}{2}a_{n-1}$$ for $n\geq2.$ Then, $\{b_n\}_{n\geq2}$ is a positive sequence satisfying $$b_n<b_{n-1}$$ for $n\geq3.$ Thus, $\{b_n\}$ is a convergent sequence. In other words, $$\lim_{n\to\infty}\left(a_n+\frac{a_{n-1}}{2}\right)=b\geq0,$$ which implies that $\{a_n\}$ is a bounded sequence, and so $$a:=\liminf_{n\to\infty}a_n\in\mathbf R\ \text{and}\ A:=\limsup_{n\to\infty}a_n\in\mathbf R.$$ Then, $$A+\frac{a}{2}=\limsup_{n\to\infty}a_n+\frac{1}{2}\liminf_{n\to\infty}a_n\leq\limsup_{n\to\infty}b_n=\liminf_{n\to\infty}b_n\leq\liminf_{n\to\infty}a_n+\frac{1}{2}\limsup_{n\to\infty}a_n=a+\frac{A}{2},$$ which gives that $A\leq a,$ and so $A=a.$
Basic Fact. For two bounded sequences $\{u_n\}$ and $\{v_n\}$, we have \begin{align*}&\liminf u_n+\liminf v_n\\\leq&\liminf(u_n+v_n)\\\leq&\liminf u_n+\limsup v_n\\\leq&\limsup(u_n+v_n)\\\leq&\limsup u_n+\limsup v_n.\end{align*}
A: It is clear that $(a_n)_{n\in\mathbb{N}}$ is bounded between $\ \max \{a_1,a_2\} $ and $0.$ Therefore by the Bolzano-Weierstrass theorem, $\ (a_n)_{n\in\mathbb{N}}\ $ has at least one limit (accumulation) point in the interval$\ [\ 0,\ \max \{a_1,a_2\}\ ].$
Now suppose, by way of contradiction, that $\ (a_n)\ $ has at least two limit points, $\ l_1\ $ and $\ l_2,\ $ with $\ l_1<l_2.\ $
Lemma: If  $\ a_{n+1} > a_n,\ $ then $\ a_k < a_n + \frac{3}{4}\left( a_{n+1} -a_n\right)\quad \forall\ k\geq n+2.$
Next, since $\ l_1\ $ and $\ l_2\ $ are limit points, $\ \exists\ N_1\ $ such that, $\ a_{N_1} < \frac{l_1+l_2}{2},\ $ and $\ a_{N_1 + 1} > \frac{l_1+l_2}{2}.\ $ The Lemma implies then that:
$$\ a_k < a_{N_1} + \frac{3}{4}\left( a_{N_1+1} -a_{N_1}\right) \leq \frac{l_1+l_2}{2} + \frac{3}{4}\left( a_{N_1+1} - \frac{l_1+l_2}{2}\right)$$
$$\implies a_k - \frac{l_1+l_2}{2} < \frac{3}{4}\left( a_{N_1+1} - \frac{l_1+l_2}{2}\right)  \quad \forall\ k\geq N_1+2.\quad (1) $$
Continuing, $\ \exists\ N_2>N_1+2\ $ such that, $\ a_{N_2} < \frac{l_1+l_2}{2},\ $ and $\ a_{N_2 + 1} > \frac{l_1+l_2}{2}.\ $ Since $\ N_2 > N_1 + 2,\ $ all $\ k\geq N_2\ $ satisfies $\ (1)\ $ also. Thus,
$$ a_k - \frac{l_1+l_2}{2} < \frac{3}{4}\left( a_{N_2+1} - \frac{l_1+l_2}{2}\right) < \frac{3}{4}\left( \frac{3}{4} \left( a_{N_1+1} - \frac{l_1+l_2}{2}\right) \right) = \left( \frac{3}{4} \right)^2 \left( a_{N_1+1} - \frac{l_1+l_2}{2}\right) $$
$$ \quad \forall\ k\geq N_2+2.\quad (2) $$
Continuing in this way, we can find and increasing sequence of integers $\ (n_j)_{j\in\mathbb{N}}\ $ such that for each $\ j,\ $ we have:
$$ a_k \leq \frac{l_1+l_2}{2} + \left(\frac{3}{4}\right)^j \left( a_{N_1 + 1} - \frac{l_1+l_2}{2} \right)\quad \forall\ k\geq N_j + 2. $$
But then, since $\  \left(\frac{3}{4}\right)^j \to 0,\ $ as $\ j\to\infty,\ $ it follows that $\ \displaystyle\limsup_k\ a_k\ \leq \frac{l_1+l_2}{2} $ contradicting the fact that $\ l_2\ $ is a limit point of $\ (a_n)_n.$
$$$$
I have left the proof of the lemma, and a couple of other details as an exercise to the reader, but these things should be relatively straightforward.
