Let $x_n = \frac{(-1)^n}{n}$. Find $\inf(T_n)$ and $\sup(T_n)$.
I believe that on this one I'm still having some trouble wrapping my head around the definition of the $nth$ tail, which I restate from my book here:
Let $x_n$ be any sequence of real numbers. Denote $T_n = \{ x_k$ | $k \geq n \}$, which we call the $nth$ tail of the sequence $x_n$.
My problem is that the $n$ in $T_n$ seems arbitrary. For example, it seems to me like,
$\sup(T_1) = \frac{1}{2}$,
$\sup(T_2) = \frac{1}{2}$,
$\sup(T_3) = \frac{1}{4}$,
$\sup(T_4) = \frac{1}{4}$,
$\sup(T_5) = \frac{1}{6}$,
$\sup(T_6) = \frac{1}{6}$,
and so on and so forth, because $T_n$ is just $x_n$, but beginning at an $n$ of our choice instead of 1. Thus, I'm not certain what it means to find the infimum and supremum of $T_n$ in general, as opposed to a particular $T_n$, where $n$ has become definite.
So, in short, what is it that I'm missing about the definition that's causing this confusion?
You can just write down your findings in terms of $n$: $$ \sup T_{n}=\sup\left\{ x_{n}\colon k\geq n\right\} =\sup\left\{ \frac{\left(-1\right)^{k}}{k}\colon k\geq n\right\} =\begin{cases} 1/n, & \text{if }n\text{ is even}\\ 1/(n+1), & \text{if }n\text{ is odd} \end{cases} $$ and similarly for the infimum.