# 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

$$\frac{11}{24}<m\le\lim_{n\rightarrow\infty}a_n\le M<\frac{23}{24}$$

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 $$\frac35\le\lim_{n\to\infty}a_n\le\frac45$$

• This is not correct. In your solution $$b_n=-\frac13b_{n-1}=\frac13\cdot\frac12b_{n-2}$$ Then $$b_n=\frac16b_{n-2}$$ This is not a minimization. For example $$b_n=\frac13b_{n-1}=\frac13\cdot\frac13b_{n-2}$$ Then $$b_n=\frac19b_{n-2}$$ Jul 15, 2017 at 14:33
• @Roman83 That does not show that my solution is incorrect. Indeed, $b_n=-\frac13\cdot-\frac12b_{n-2}$. Be wary of signs. Jul 15, 2017 at 14:36
• For the infinite sum $$b_0=1, b_1=-\frac13, b_2=\frac16, b_3=-\frac1{3\cdot2\cdot3}, ....$$ So? Jul 15, 2017 at 14:42
• @Roman83 Minimising the sum must start with $b_1=-\frac12$, NOT $b_1=-\frac13$! Jul 15, 2017 at 14:43
• "question eligible for bounty in 23 hours" Jul 15, 2017 at 15:43

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.

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 $4/5$.