6
$\begingroup$

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

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

$\endgroup$
5
+150
$\begingroup$

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

$\endgroup$
14
  • $\begingroup$ 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 $$b_n=\frac19b_{n-2}$$ $\endgroup$
    – Roman83
    Jul 15 '17 at 14:33
  • $\begingroup$ @Roman83 That does not show that my solution is incorrect. Indeed, $b_n=-\frac13\cdot-\frac12b_{n-2}$. Be wary of signs. $\endgroup$ Jul 15 '17 at 14:36
  • $\begingroup$ For the infinite sum $$b_0=1, b_1=-\frac13, b_2=\frac16, b_3=-\frac1{3\cdot2\cdot3}, ....$$ So? $\endgroup$
    – Roman83
    Jul 15 '17 at 14:42
  • $\begingroup$ @Roman83 Minimising the sum must start with $b_1=-\frac12$, NOT $b_1=-\frac13$! $\endgroup$ Jul 15 '17 at 14:43
  • 1
    $\begingroup$ "question eligible for bounty in 23 hours" $\endgroup$
    – Roman83
    Jul 15 '17 at 15:43
4
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $4/5$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.