Convergent sequence with odd terms decreasing and even terms increasing 
Let $\left(a_n\right)$ is convergent sequence. $a_0=0, a_1=1,a_2,a_3,...$
Odd terms decrease and even terms increase and for all $n\ge1$ 
  $$2\le \frac{a_n-a_{n-1}}{a_n-a_{n+1}}\le3.$$
  Find the boundaries in which there can be a limit of this sequence.

My work so far:
I proved that $$\frac{11}{24}\le\lim_{n\to\infty}a_n\le\frac{23}{24}$$
But I can not say that this is the final answer. I need an examples of sequences such that:
1) $$\lim_{n\rightarrow\infty}a_n=\frac{11}{24}$$
2) $$\lim_{n\rightarrow\infty}a_n=\frac{23}{24}$$
or 
3) (If my answer can be improved) I need numbers $m$ and $M$, where 
$$\frac{11}{24}<m\le\lim_{n\rightarrow\infty}a_n\le M<\frac{23}{24}$$ 
 A: Define the first differences $b_n=a_{n+1}-a_n$ for $n\ge0$. Then $b_0=1$, and since $a_0=0$:
$$\lim_{n\to\infty}a_n=\sum_{n=0}^\infty b_n$$
The given inequality may be rewritten for $n\ge0$ as
$$-\frac12\le\frac{b_{n+1}}{b_n}\le-\frac13$$
To make the infinite sum in $b_n$ as large as possible we have to


*

*Subtract as little as possible: for odd-numbered $b_n$, which will be negative, we multiply by $-\frac13$ from $b_{n-1}$

*Add as much as possible: for even-numbered $b_n$, which will be positive, we multiply by $-\frac12$ from $b_{n-1}$


This results in
$$S=1-\frac13+\frac1{3\cdot2}-\frac1{3\cdot2\cdot3}+\dots$$
Multiply both sides by $\frac1{3\cdot2}$:
$$\frac16S=\frac1{3\cdot2}-\frac1{3\cdot2\cdot3}+\dots=S-1+\frac13$$
$$-\frac56S=-\frac23\qquad S=\frac45$$
Similarly we can make the infinite sum in $b_n$ as small as possible by swapping $-\frac13$ and $-\frac12$ in the list above, giving
$$T=1-\frac12+\frac1{2\cdot3}-\frac1{2\cdot3\cdot2}+\dots$$
Once again, multiply both sides by $\frac1{2\cdot3}$:
$$\frac16T=\frac1{2\cdot3}-\frac1{2\cdot3\cdot2}+\dots=T-1+\frac12$$
$$-\frac56T=-\frac12\qquad T=\frac35$$
Therefore
$$\frac35\le\lim_{n\to\infty}a_n\le\frac45$$
A: The answer of Parcly is correct and should be awarded the bounty. For a formal proof: Let $B$ denote the set of admissible sequences $\beta=(b_0=1,b_1,b_2,...)$ verifying that $b_{2n}>0>b_{2n+1}$ all $n$ and the ratio condition
  $$  b_k/b_{k+1} \in \Delta=\left[-\frac12,-\frac13\right]$$
We are looking for extremal values of $\sum_{k\geq 0} b_k$.
The key point is that for any $m\geq 0$
 one has the following a priori bound (split into even and odd indices):
$$ (-1)^m \sum_{k\geq 0} b_{m+k} \geq 
(-1)^m b_m \left( \sum_{k\geq 0} (1/9)^k - 1/2 \sum_{k\geq 0} (1/4)^k\right) =(-1)^m b_m \frac{11}{24}>0$$
in particular,  the tail-sum from term $m$ has the same sign as $b_m$.
Suppose now that $\beta$ is admissible then for $m\geq 1$ so is 
$$\hat{\beta}_{m,r} = (b_0=1,...,b_{m-1}, r b_{m}, r b_{m+1},...)$$
for any $r>0$ for which  $rb_m/b_{m+1}\in \Delta$.
The sum of the series $\hat{\beta}_{m,r}$ is $\sum_{0\leq k<m} b_k + r \sum_{k\geq m} b_k$. As shown above the last sum has the same sign as $b_m$ 
and can be made strictly smaller and larger by choosing suitable $r$ whenever $b_m/b_{m+1}\in (-\frac12,-\frac13) $  (an interior point). For example to minimize the sum, for every even $m$ we must require $b_m/b_{m-1}=-1/3$ or else  you may make the sum of $\hat{\beta}_{m,r}$ smaller for suitable $r<1$. Similarly for odd $m$, we must have $b_m/b_{m-1}=-1/2$.
The max case is treated in a similar way and the extremal values are then given as described by Parcly.
A: I will consider the general case of a sequence
$(a_n)_{n\ge0}$ such that $a_0=0$ $a_1=1$, the subsequence $(a_{2n})_{n\ge0}$ is increasing, and
    the sequence $(a_{2n+1})_{n\ge0}$ is decreasing, and finally for $0< \beta<\alpha<1$ we have
    $$
 \forall\, n\ge1,\qquad
 \frac{1}{\alpha} \le \frac{a_n-a_{n-1}}{a_n-a_{n+1}}\le\frac{1}{\beta}\tag{$*$}$$
    I will prove that $\lim\limits_{n\to\infty}a_n$ exists and that    $$\frac{1-\alpha}{1-\alpha\beta}\le\lim_{n\to\infty}a_n\le
\frac{1-\beta}{1-\alpha\beta}$$
and finally that this conclusion cannot be improved.
Let $b_n=(-1)^n(a_{n+1}-a_{n})$. Then $(*)$ implies that $b_{n-1}/b_n\ge\alpha>0$ so all the terms of the sequence $(b_n)_{n\ge0}$ have the same sign, but
$b_0=1>0$, hence $b_n>0$ for all $n$. Moreover, $(*)$ implies also that
$b_{n+1}\le\alpha b_n$ for all $n\ge0$, thus $b_n\le\alpha^{n}$ and the
the series $\sum_{n=0}^\infty (a_{n+1}-a_{n})$ is absolutely convergent, which is equivalent to the existence of  $\ell=\lim\limits_{n\to\infty}a_n$.
Now, considering two cases $(*)$ is equivalent to the following two inequalities:
\begin{alignat*}{3}
&(1-\alpha)a_{2n+1}+\alpha a_{2n}&&\le a_{2n+2}&&\le
 (1-\beta)a_{2n+1}+\beta a_{2n}\tag{1}\\
&(1-\beta) a_{2n+2}+ \beta a_{2n+1}&&\le a_{2n+3}&&\le (1-\alpha)a_{2n+2}+\alpha a_{2n+1}\tag{2}
\end{alignat*}
This suggests that we consider the sequences $(m_n)_{n\ge0}$ and $(M_n)_{n\ge0}$ defined as follows:
\begin{alignat*}{3}
&m_0=0,m_1=1, \quad
m_{2n+2}&&=(1-\alpha)m_{2n+1}+\alpha m_{2n},\quad m_{2n+3}&&=(1-\beta)m_{2n+2}+\beta m_{2n+1}.\\
&M_0=0,M_1=1, \quad
M_{2n+2}&&=(1-\beta)M_{2n+1}+\beta M_{2n},\quad M_{2n+3}&&=(1-\alpha)M_{2n+2}+\alpha M_{2n+1}.
\end{alignat*}
Then it is an easy induction to prove that
\begin{equation*}
\forall\,n\ge0,\quad m_n\le a_n\le M_n\tag{3}
\end{equation*}
Indeed, let $\mathbb{P}_n=(
m_{n}\le a_{n}\le M_{n})$, then the base cases $\mathbb{P}_0$ and $\mathbb{P}_1$ are satisfied by assumption. Now, from $(1)$ we conclude that
$(\mathbb{P}_{2n}\wedge \mathbb{P}_{2n+1})\implies \mathbb{P}_{2n+2}$ and
from $(2)$ we conclude that
$(\mathbb{P}_{2n+1}\wedge \mathbb{P}_{2n+2})\implies \mathbb{P}_{2n+3}$
This proves that $\mathbb{P}_n$ is satisfied for every $n$.
Now, if $X_n=\left[\begin{matrix}
m_{2n}\\ m_{2n+1}
\end{matrix}\right]$, and $Y_n=\left[\begin{matrix}
M_{2n}\\ M_{2n+1}
\end{matrix}\right]$ then
\begin{equation*}
X_{n+1}=\underbrace{\left[\begin{matrix}
\alpha&1-\alpha\\
(1-\beta)\alpha&1-\alpha+\alpha\beta
\end{matrix}\right]}_{A_{\alpha,\beta}}X_n,\qquad
Y_{n+1}=\underbrace{\left[\begin{matrix}
\beta&1-\beta\\
(1-\alpha)\beta&1-\beta+\alpha\beta
\end{matrix}\right]}_{A_{\beta,\alpha}}Y_n
\end{equation*}
with initial conditions $X_0=Y_0=\left[\begin{matrix}
0\\1
\end{matrix}\right]$.
Now, the characteristic polynomial $Q(X)$ of $A_{\alpha,\beta}$ is given by
$Q(X)=X^2-(1+\alpha\beta)X+\alpha\beta=(X-1)(X-\alpha\beta)$, and it is easy to check that the remainder of the euclidean division of $X^n$ by $Q(X)$ is
\begin{equation*}
\frac{1}{1-\alpha\beta}(X-\alpha\beta)+
\frac{(\alpha\beta)^n}{1-\alpha\beta}(1-X)
\end{equation*}
Hence
\begin{equation*}
A_{\alpha,\beta}^n=
\frac{1}{1-\alpha\beta}(A_{\alpha,\beta}-\alpha\beta I)
+\frac{(\alpha\beta)^n}{1-\alpha\beta}(I-A_{\alpha,\beta})
\end{equation*}
and since $X_n=A_{\alpha,\beta}^nX_0$ we conclude easily that
\begin{align*}
m_{2n}&=\frac{1-\alpha}{1-\alpha\beta}-\frac{1-\alpha}{1-\alpha\beta}(\alpha\beta)^n\\
m_{2n+1}&=\frac{1-\alpha}{1-\alpha\beta}+\frac{\alpha(1-\beta)}{1-\alpha\beta}(\alpha\beta)^n
\end{align*}
Exchanging the roles of $\alpha$ and $\beta$ we see also that
\begin{align*}
M_{2n}&=\frac{1-\beta}{1-\alpha\beta}-\frac{1-\beta}{1-\alpha\beta}(\alpha\beta)^n\\
M_{2n+1}&=\frac{1-\beta}{1-\alpha\beta}+\frac{\beta(1-\alpha)}{1-\alpha\beta}(\alpha\beta)^n
\end{align*}
In particular,
we have \begin{equation*}
\lim_{n\to\infty}m_n=\frac{1-\alpha}{1-\alpha\beta},
\quad \lim_{n\to\infty}M_n=\frac{1-\beta}{1-\alpha\beta}
\end{equation*}
Hence, letting $n$ tend to $+\infty$ in $(3)$ we get
\begin{equation*}
\frac{1-\alpha}{1-\alpha\beta}\le\lim_{n\to\infty}a_n\le
\frac{1-\beta}{1-\alpha\beta}
\end{equation*} 
Now, taking $a_n=m_n$ for all $n$, shows that the lower bound in the above inequality is the best possible, because it is attained, and taking $a_n=M_n$ for all $n$, shows that the upper bound in the above inequality is also the best possible, because it is attained.
Further, considering sequences $(a_n)_{n\ge0}$
 of the form $a_n=tm_n+(1-t)M_n$ where $0<t<1$, shows that for any number $\ell$ in the interval $[\frac{1-\alpha}{1-\alpha\beta},
 \frac{1-\beta}{1-\alpha\beta}]$ there exists a sequence  $(a_n)_{n\ge0}$
 satisfying the conditions of the problem and converging to $\ell$.
Remark. Surly we have noticed that the proposed problem corresponds to the case $\alpha=1/2$ and $\beta=1/3$, so the lower and upper bounds are indeed $3/5$ and $4/5$.
