MHB Help with finding the sup and inf

[Saitama]

Problem:
Let $\displaystyle S=\left\{1-\frac{(-1)^n}{n}:n\in \mathbb{N}\right\}$. Find $\inf S$ and $\sup S$.

Attempt:
I know the problem is quite easy and the answers are very obvious but I need help with writing down a proof for it and I am really bad at writing such proofs.

The best I could come up with was to divide the given set into two subsets. First set comprises of elements (I am not sure if this is the right word) when $n=2k$ ($S_1$) and the other for $n=2k-1$ ($S_{2n-1}$) for $k\in\mathbb{N}$. i.e
$$S_1=\left\{1-\frac{1}{2k}:k\in \mathbb{N}\right\}$$
and
$$S_2=\left\{1+\frac{1}{2k-1}:k\in\mathbb{N}\right\}$$
Now it is obvious that $\sup S$ is in $S_2$ and $\inf S$ is in $S_1$. It is easy to see that both $\sup S$ and $\inf S$ occur when $k=1$.

But how do I write down a "proof"?

Any help is appreciated. Thanks!

 

[I like Serena]

Hi Pranav!

I would write:

Proof
$$\forall n \in \mathbb N: 1-\frac{(-1)^2}{2} \le 1 - \frac{(-1)^n}{n} \le 1 - \frac{(-1)^1}{1}$$
Equalities are achieved for $n=1$ respectively $n=2$.
Therefore $\inf S = \frac 12$ and $\sup S = 2$.

 

[Prove It]

Problem:
Let $\displaystyle S=\left\{1-\frac{(-1)^n}{n}:n\in \mathbb{N}\right\}$. Find $\inf S$ and $\sup S$.

Attempt:
I know the problem is quite easy and the answers are very obvious but I need help with writing down a proof for it and I am really bad at writing such proofs.

The best I could come up with was to divide the given set into two subsets. First set comprises of elements (I am not sure if this is the right word) when $n=2k$ ($S_1$) and the other for $n=2k-1$ ($S_{2n-1}$) for $k\in\mathbb{N}$. i.e
$$S_1=\left\{1-\frac{1}{2k}:k\in \mathbb{N}\right\}$$
and
$$S_2=\left\{1+\frac{1}{2k-1}:k\in\mathbb{N}\right\}$$
Now it is obvious that $\sup S$ is in $S_2$ and $\inf S$ is in $S_1$. It is easy to see that both $\sup S$ and $\inf S$ occur when $k=1$.

But how do I write down a "proof"?

Any help is appreciated. Thanks!

For starters, sometimes a supremum is called the "least upper bound", which is the smallest value that can be used as a ceiling for your sequence, and sometimes an infemum is called the "greatest lower bound", which is the largest value that can be used as the floor.

For starters, I would write $\displaystyle \begin{align*} 1 - \frac{ \left( -1 \right) ^n }{n} = 1 + \frac{\left( -1 \right) ^{n+1}}{n} \end{align*}$

I also like to write down a few terms to determine the behaviour of the function. To do this, a start is to write down the first few terms of the sequence. For simplicity I'll leave out the +1...

$\displaystyle \begin{align*} \left\{ \frac{ \left( -1 \right) ^{n + 1}}{n} \right\} = \left\{ 1, -\frac{1}{2}, \frac{1}{3}, -\frac{1}{4}, \dots \right\} \end{align*}$

It should be obvious that the terms decrease in magnitude (mainly because $\displaystyle \begin{align*} \frac{1}{n + 1} < \frac{1}{n} \end{align*}$. Thus the supremum of this is 1 and the infemum is $\displaystyle \begin{align*} -\frac{1}{2} \end{align*}$.

So for your sequence $\displaystyle \begin{align*} \left\{ 1 + \frac{ \left( -1 \right) ^{n + 1}}{n} \right\} \end{align*}$ has a supremum of 2 and an infemum of $\displaystyle \begin{align*} \frac{1}{2} \end{align*}$.