# Thread: Maximum, Minimum, Supremum and Infimum

1. ## Maximum, Minimum, Supremum and Infimum

please explain me how i'm solving it and how i should present the solution :

thanks!

2. In a), I think you could solve it by looking at the function

$F(k)=\prod_{n=1}^k\left(1+\frac{1}{n}\right)^{1/k} = e^{f(k)}$

where

$f(k)=\ln\left(\prod_{n=1}^k\left(1+\frac{1}{n}\rig ht)^{1/k}\right) = \sum_{n=1}^k\frac{\ln\left(1+\cfrac{1}{n}\right)}{ k}$

and by realizing that a k that gives a minimal or a maximal value for f, will also give a minimal or a maximal value, respectively, for F.

Also note that

$\ln\left(1+\cfrac{1}{n}\right) \leq\ \frac{1}{n}$

3. I'm going to assume that I am interpreting these questions correctly.

Problem: Let $\mathcal{I}=\left\{x: x=\left[\left(1+\frac{1}{1}\right)\cdots\left(1+\frac{1}{k }\right)\right]^{\frac{1}{k}}\quad k\in\mathbb{N}\right\}$. Find $\max\text{ }\mathcal{I},\min\text{ }\mathcal{I},\inf\text{ }\mathcal{I},\sup\text{ }\mathcal{I}$

Solution: There is no "minimum" of this set. Those concepts (at least for how I have them defined) act only on members of the set. So $\max\text{ }\mathcal{I}$ existing would imply $\exists \delta\in\mathcal{I}$ such that $\forall x\in\mathcal{I}\quad x\le\delta$. Before we proceed we need a lemma.

Lemma: Let $\sigma_k=\ln\left\{\left[\left(1+\frac{1}{1}\right)\cdots\left(1+\frac{1}{k }\right)\right]^{\frac{1}{k}}\right\}=\frac{1}{k}\sum_{\ell=1}^k \ln\left[1+\frac{1}{\ell}\right]$, then $\sigma_{k+1}<\sigma_{k}\quad\forall k\in\mathbb{N}$

Proof: This is just true by a series of observations. Namely:

$\ln\left[1+\frac{1}{k+1}\right]\le \frac{1}{k}\sum_{\ell=1}^{k}\ln\left[1+\frac{1}{\ell}\right]$. Adding $\sum_{\ell=1}^{k}\ln\left[1+\frac{1}{\ell}\right]$ to both sides gives $\sum_{\ell=1}^{k}\ln\left[1+\frac{1}{\ell}\right]+\ln\left[1+\frac{1}{k+1}\right]=\sum_{\ell=1}^{k+1}\ln\left[1+\frac{1}{\ell}\right]\le\left(1+\frac{1}{k}\right)\sum_{\ell=1}^{k}\ln\ left[1+\frac{1}{\ell}\right]$. Dividing both sides by $k+1$ gives $\frac{1}{k+1}\sum_{\ell=1}^{k+1}\ln\left[1+\frac{1}{\ell}\right]\le\frac{1}{k+1}\left(1+\frac{1}{k}\right)\sum_{\e ll=1}^k\ln\left[1+\frac{1}{\ell}\right]=\frac{1}{k}\sum_{\ell=1}^{k}\ln\left[1+\frac{1}{\ell}\right]$ The conclusion follows. $\blacksquare$

Since $\ln(x)$ is an increasing function this implies that $\left[\left(1+\frac{1}{1}\right)\cdots\left(1+\frac{1}{k }\right)\right]^{\frac{1}{k}}$ is a decreasing function.

Thus we can show that $\min\text{ }\mathcal{I}$ does not exist. We do this by contradiction. Suppose $\max\text{ }\mathcal{I}=\delta$. Then $\delta=\left[\left(1+\frac{1}{1}\right)\cdots\left(1+\frac{1}{k }\right)\right]^{\frac{1}{k}}$ for some $k\in\mathbb{N}$. But $\left[\left(1+\frac{1}{1}\right)\cdots\left(1+\frac{1}{k +1}\right)\right]^{\frac{1}{k+1}}\le \delta$ and $\left[\left(1+\frac{1}{1}\right)\cdots\left(1+\frac{1}{k +1}\right)\right]^{\frac{1}{k+1}}\in\mathcal{I}$ contradicting our choice of $\delta$. Therefore $\min\text{ }\mathcal{I}$ does not exist.

From our lemma we can easily deduce that $2\in\mathcal{I}$ and $\delta\le2\quad\forall \delta\in\mathcal{I}$. Thus, $\max\text{ }\mathcal{I}=2$. Consequently, $\sup\text{ }\mathcal{I}=2$

Lastly we must find $\inf\text{ }\mathcal{I}$. We claim that it is.......I leave the rest for you.

4. wow!! i can't even understand what you wrote, it suppose to be more simple, we didn't learn that, i'm sure.

5. Well, you should have learned to at least calculate some of those numbers!

In the first problem, you have numbers of the form $\left[\left(1+\frac{1}{1}\right)\cdots\left(1+\frac{1}{k }\right)\right]^{\frac{1}{k}}$.

When k= 1, that is $(1+ 1)^{1/1}= 2$.

When k= 2, that is $[(1+1/1)(1+1/2)]^{1/2}= [(2)(3/2)]^{1/2}= 3^{1/2}$.

When k= 3, that is $[(1+1/1)((1+ 1/2)(1+ 1/3)]^{1/3}= [(3)(4/3)]^{1/3}= 4^{1/3}$.

That is, $(1+ 1/1)(1+ 1/2)(1+1/3)\cdot\cdot\cdot(1+ 1/(k-1))(1+1/k)$ $= (2)(3/2)(4/3)\cdot\cdot\cdot k/(k-1))((k+1)/k= k+1$ so your sequence reduces to $(k+1)^{1/k}$. It should be easy to see that that is a decreasing sequence.

6. Originally Posted by HallsofIvy
That is, $(1+ 1/1)(1+ 1/2)(1+1/3)\cdot\cdot\cdot(1+ 1/(k-1))(1+1/k)=$ $(2)(3/2)(4/3)\cdot\cdot\cdot(k/(k-1))((k+1)/k)= k+1$ so you sequence reduces to $(k+1)^{1/k}$. It should be easy to see that that is a decreasing sequence.
Nice! I got to try harder on seeing those things.