# Unbounded sequence?

• Oct 4th 2008, 02:50 AM
bkarpuz
Unbounded sequence?
I have a solution for the following problem but I really wonder what kind of ways will be provided for this.

Problem. Let
$\displaystyle a(m,n)= \begin{cases} 67/256,&\mathrm{mod}(m,4)=0\text{ and }\mathrm{mod}(n,2)=0\\ %\delta 27/256,&\text{ otherwise}. \end{cases}$
Prove that
$\displaystyle \limsup\limits_{m,n\to\infty}A(m,n)=\infty,$
where
$\displaystyle A(m,n):=\sum\limits_{i=1}^{m}\sum\limits_{j=1}^{n} a(i,j)\Bigg\{\frac{4}{3}\Bigg(\sum\limits_{u=i+1}^ {i+3}a(u,j+1)\Bigg)^{1/4}-1\Bigg\}$
.................$\displaystyle -\frac{1}{3}\Bigg[\sum\limits_{i=m-3}^{m}\sum\limits_{j=1}^{n}\sum\limits_{u=i+1}^{i+ 3}a(u,j+1)+\sum\limits_{i=1}^{m}\sum\limits_{j=n-1}^{n}\sum\limits_{u=i+1}^{i+3}a(u,j+1)\Bigg].$
• Oct 4th 2008, 04:04 AM
Laurent
The sequence $\displaystyle a(m,n)$ is "almost" constant. Let's bound $\displaystyle A(m,n)$ from below by replacing $\displaystyle a(m,n)$ by $\displaystyle a=\frac{27}{256}$ in the first term and by $\displaystyle a'=\frac{67}{256}$ in the second one (after the minus sign): we get $\displaystyle A(m,n)\geq mna(\frac{4}{3}(3a)^{1/4}-1)-(12n+6m)a'$. However, we have $\displaystyle \frac{4}{3}(3a)^{1/4}=\left(\frac{256 a}{27}\right)^{1/4}=1$ so that this lower bound is slightly too strong, but if we had a positive factor instead in front of $\displaystyle mn$ (no matter how small), this would be right. Then, to bound the first term, it suffices to consider the indices where $\displaystyle 4|i+1$ and $\displaystyle 2|j+1$ (you can bound the terms with other indices by 0 thanks to the previous computation). There are about $\displaystyle \frac{mn}{8}$ such couples $\displaystyle (i,j)$, and for each of them the summand is greater than $\displaystyle a(\frac{4}{3}(2a+a')^{1/4}-1)$, which is strictly greater than $\displaystyle a(\frac{4}{3}(3a)^{1/4}-1)=0$. We get that $\displaystyle A(m,n)\geq \lambda mn- \alpha m - \beta n$ for some positive $\displaystyle \lambda, \alpha,\beta$. Choosing $\displaystyle m=n$ for instance, you easily see that this is unbounded from above.
• Oct 4th 2008, 04:35 AM
bkarpuz
Here is my solution.
Solution. Clearly, we have $\displaystyle a(m+4,n)=a(m,n)$ and $\displaystyle a(m,n+2)=a(m,n)$ for all $\displaystyle m,n\in\mathbb{N}$.
Therefore, we have
$\displaystyle \sum\limits_{i=m-3}^{m}\sum\limits_{j=n-1}^{n}a(i,j)\Bigg\{\frac{4}{3}\Bigg(\sum\limits_{u =i+1}^{i+3}a(u,j+1)\Bigg)^{1/4}-1\Bigg\}=\frac{81}{256}\bigg(\frac{\sqrt{11}}{3}-1\bigg)>0$........(1)
and
$\displaystyle \sum\limits_{i=m-3}^{m}\sum\limits_{j=n-1}^{n}\sum\limits_{u=i+1}^{i+3}a(u,j+1)=3$.................................................. ...............(2)
for all $\displaystyle m,n\in\mathbb{N}$.
Therefore, using (1) and (2), we have the following inequalities:
$\displaystyle \sum\limits_{i=1}^{m}\sum\limits_{j=1}^{n}a(i,j)\B igg\{\frac{4}{3}\Bigg(\sum\limits_{u=i+1}^{i+3}a(u ,j+1)\Bigg)^{1/4}-1\Bigg\}\geq\frac{81}{256}\bigg(\frac{\sqrt{11}}{3 }-1\bigg)\bigg\lfloor\frac{m}{4}\bigg\rfloor\bigg\lf loor\frac{n}{2}\bigg\rfloor$......(3)
$\displaystyle \sum\limits_{i=m-3}^{m}\sum\limits_{j=1}^{n}\sum\limits_{u=i+1}^{i+ 3}a(u,j+1)\leq3\bigg\lceil\frac{n}{2}\bigg\rceil$.................................................. .........._(4)
$\displaystyle \sum\limits_{i=1}^{m}\sum\limits_{j=n-1}^{n}\sum\limits_{u=i+1}^{i+3}a(u,j+1)\leq3\bigg\ lceil\frac{m}{4}\bigg\rceil$.................................................. ..........,(5)
for all $\displaystyle m,n\in\mathbb{N}$.
Hence, from (3)--(5), we learn that
$\displaystyle A(m,n)\geq B(m,n)$
holds for all $\displaystyle m,n\in\mathbb{N}$, where
$\displaystyle B(m,n):=\frac{81}{256}\bigg(\frac{\sqrt{11}}{3}-1\bigg)\bigg\lfloor\frac{m}{4}\bigg\rfloor\bigg\lf loor\frac{n}{2}\bigg\rfloor-\Bigg\{\bigg\lceil\frac{n}{2}\bigg\rceil+\bigg\lce il\frac{m}{4}\bigg\rceil\Bigg\}$
for $\displaystyle m,n\in\mathbb{N}$.
It suffices to prove that $\displaystyle B$ is unbounded.
As $\displaystyle k\to\infty$, we have
$\displaystyle B(4k,2k)=\frac{81}{256}\bigg(\frac{\sqrt{11}}{3}-1\bigg)k^{2}-2k\to\infty$.
Thus, the proof is done. $\displaystyle \rule{0.3cm}{0.3cm}$

Notation. $\displaystyle \lfloor\cdot\rfloor$ denotes the least integer function and $\displaystyle \lceil\cdot\rceil$ denotes the greatest integer function.

Appendix. Here, (2) is shown.
By the periodicity, we obtain
$\displaystyle \sum\limits_{i=m-3}^{m}\sum\limits_{j=n-1}^{n}a(i,j+1)=1$.................................................. .......................(6)
for all $\displaystyle m,n\in\mathbb{N}$.
Using (6), we get
$\displaystyle \sum\limits_{i=m-3}^{m}\sum\limits_{j=n-1}^{n}\sum\limits_{u=i+1}^{i+3}a(u,j+1)=\sum\limit s_{i=m-3}^{m}\sum\limits_{u=i+1}^{i+3}\sum\limits_{j=n-1}^{n}a(u,j+1)$
......................................_$\displaystyle =\sum\limits_{u=1}^{3}\sum\limits_{i=m+u-3}^{m+u}\sum\limits_{j=n-1}^{n}a(i,j+1)$
......................................_$\displaystyle =\sum\limits_{u=1}^{3}1=3$
for all $\displaystyle m,n\in\mathbb{N}$.