Supremum and infimum of a set of reals Find the supremum and infimum of the following set $$ A:= \left\\{1 + \frac{(-1)^n}{n} \mid n\in \mathbb{N}\right\\}.$$ Here if $n$ is near infinity then the fraction part is $0$, so the value of $A$ is $1$. So any of the terminal values are is $1$. It can be supremum or infimum. But Textbook says that the supremum is $\frac{3}{2}$. And the infimum is $0$. Can anyone show me how?

For a set $A=\\{a_n~:~n\in\mathbb N\\}$ the supremum and infimum is not the same as the limes superior or limes inferior of the sequence $(a_n)_n$. You might compute a few elements \begin{align} A&=\left\\{1-\frac11,1+\frac12,1-\frac13,1+\frac14,1-\frac15,1+\frac16,1-\frac17\ldots\right\\}\\\ &=\left\\{0,\frac32,\frac23,\frac54,\frac45,\frac76,\frac67,\ldots\right\\} \end{align} The supremum of $A$ is the smallest upper bound of $A$. You can see that it might be $\frac32$ since it seems that all other elements are less then $\frac32$. But you need to prove it. This is given by $$ \frac32=1+\frac12\geq 1+\frac1n\geq 1+\frac{(-1)^n}{n} \text{ for all }n\geq 2 $$ and $\frac32>0=1+\frac{(-1)^1}1$. Therefore $\frac32$ is an upper bound of $A$ and since $\frac32\in A$ it is the smallest upper bound.

Same way you can argue why $0$ is the biggest lower bound of $A$.