I have a question about bounds and others.
So, question is that: $D$ is the set $D = \bigl\{ \frac{(-1)^n n}{n+1}\colon n \in \Bbb{N} \bigr\}$.
How to find $\inf(D)$ and $\sup(D)$ and show they do not belong to $D$?
My approach is that: Clearly, $1$ is an upper bound for $D$. Let $M$ be the upper bound that is less than $1$.
I know that my approach is nothing to talk about but this is all I had. Please help me.
You are on the right track. note that the same argument for sup and inf will work, since this alternates. The key is to realize that $\frac n {n+1}$ gets as close to 1 as you want. The easy way to see this is to do the following rewriting:
$$\frac n {n+1}=\frac {n +1 -1} {n+1}=\frac {n+1}{n+1}-\frac 1 {n+1}=1-\frac 1 {n+1}$$
so the positive terms get as close to 1 as you want. The formal rule for this is called the Archimedean property of the real numbers, for any $\epsilon>0$ we can find a natural number $n$ such that $\frac 1 n <\epsilon$.
So, if you had an upper bound that was smaller than 1, call it D, we would have $D<1$, or $0<1-D$. Thus, by the Archimedean property, there is some natural number $n$ such that $\frac 1 n < 1 -D$.
Adding $D$ and subtracting $\frac 1 n$ gets us to $$D<1-\frac 1 n$$, but that contradicts $D$ being an upper bound. (Technically we have to restrict $n$ to be even due to the alternating nature of the sequence to get the positive side, but that's no issue, because once you have it works for one $n$, it works for all bigger $n$'s.
Can you mirror this argument for the infimum?