Least upper and greatest lower bounds

Question

Let $n\ge 3$ be an arbitrarily fixed integer. Take all the possible finite sequences $(a_1,\dots,a_n)$ of positive numbers. Find the least upper and the greatest lower bounds of the set of numbers $$ \sum_{k=1}^n \dfrac{a_k}{a_k+a_{k+1}+a_{k+2}} $$ where we put $a_{n+1}=a_1$ and $a_{n+2}=a_2$.

I've been struggling to solve this problem from the book "Problems in Mathematical Analysis I" from W.J. Kaczor and M.T. Nowak but I've given up. As the book provides a solution I've checked it, but there's something I don't understand:

When it says $\textit{To this end we take } a_k=t^k,t>0$, what result is this? Where can I find it?

Another question: how to get $$ \dfrac{a_k}{a_k+a_{k+1}+a_{k+2}}\le 1-\dfrac{a_{k+1}}{s}-\dfrac{a_{k+2}}{s} $$ I guess it's using the fact that $$ \dfrac{a}{b}\le \dfrac{c}{d} \implies \dfrac{a}{b}\le \dfrac{a+c}{b+d} $$ is my assumption right? If not I don't see how to obtain the upper bound either.

Thanks in advance.

Answer 1

1) He's not saying "Let's take a previous result where we proved $a_k = t^k$". He's saying "Let's take the sequence where $a_k:= t^k$". It is just one possible sequence.

2) $s = a_1 + ........ + a_n$ so

$\dfrac {a_k}s = \dfrac {a_k}{a_1 + a_2 + .... + a_n} \le \dfrac{a_k}{a_k+a_{k+1}+a_{k+2}}$

because $s \ge a_k+a_{k+1}+a_{k+2}$

$\dfrac{a_k}{a_k+a_{k+1}+a_{k+2}}= \dfrac {a_k+a_{k+1}+a_{k+2} -a_{k+1} - a_{k+2}}{a_k+a_{k+1}+a_{k+2}}=$

$ 1 - \dfrac{a_{k+1}}{a_k+a_{k+1}+a_{k+2}}- \dfrac{a_{k+2}}{a_k+a_{k+1}+a_{k+2}}\le$

$ 1 - \dfrac{a_{k+1}}{s}- \dfrac{a_{k+2}}{s}$

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Outline of his proof.

1) Prove for all $\{a_k\}$ that $\dfrac{a_k}{a_k+a_{k+1}+a_{k+2}} \ge 1$. So $1$ is a lower bound.

2) Prove for all $\{a_k\}$ that $\dfrac{a_k}{a_k+a_{k+1}+a_{k+2}} \le n-2$. So $n-2$ is an upper bound.

3) Show that if we take the sequences $\{a_{k,t}\} = \{t^k\}$ then

3a) $$\lim\limits_{t\to 0} \sum \dfrac{a_{k,t}}{a_{k,t}+a_{k+1,t}+a_{k+2,t}}=\lim\limits_{t\to 0}[ \sum_{k=1}^{n-2}\dfrac{t^k}{t^k+t^{k+1}+t^{k+2}} + \dfrac{t^{n-1}}{t^{n-1}+t^{n}+t} + \dfrac{t^{n}}{t^{n}+t+t^2}]=1$$ so $1$ is not just a lower bound it is the greatest lower bound.

3b)$$\lim\limits_{t\to \infty} \sum \dfrac{a_{k,t}}{a_{k,t}+a_{k+1,t}+a_{k+2,t}}=\lim\limits_{t\to \infty}[ \sum_{k=1}^{n-2}\dfrac{t^k}{t^k+t^{k+1}+t^{k+2}} + \dfrac{t^{n-1}}{t^{n-1}+t^{n}+t} + \dfrac{t^{n}}{t^{n}+t+t^2}]=n-2$$ so $n-2$ is not just an upper bound it is the least lower bound.

Answer 2

1.) That's not a result, that's just a certainly allowed way to pick specific values of the $a_n$ (just with less of those pesky degrees of freedom).

2.) No. Simply apply $\sum_{k=1}^n$ to $$\frac{a_k}s\le \frac{a_k}{a_k+a_{k+1}+a_{k+2}}\le1-\frac{a_{k+1}}2-\frac{a_{k+1}}s $$ to obtain $$1=\frac ss=\frac{\sum a_k}s\le \sum\frac{a_k}{a_k+a_{k+1}+a_{k+2}}\le\sum1-\frac{\sum a_{k+1}}2-\frac{\sum a_{k+1}}s=n-2 $$