For which values of $\alpha$ is {$z_n$} a bounded sequence? Where $\alpha$ is a real constant, consider the sequence {$z_n$} defined by $z_n=\frac{1}{n^\alpha}$. For which value of $\alpha$ is {$z_n$} a bounded sequence?
How do I start with this kind of question? I think that $\forall\space \alpha\in\Bbb{R}_{\geq0}$  the sequence is convergent and therefore bounded, but how do I write it out?
 A: If $\alpha=0$, $(z_n)$ is constant, hence bounded.
If $\alpha>0$, $(z_n)$ converges to 0 and is thus bounded.
If $\alpha<0$, $(z_n)$ diverges to $+\infty$ and is thus unbounded.
A: As I state in the comment, you have the correct answer. The only remaining task is to give a formal explanation of the answer.  One way to write an answer is as follows:
First, we note that the function $f: \Bbb [1,\infty) \to \Bbb R$ defined by $f(x) = x^{\beta}$ satisfies
$$
\lim_{x \to \infty}f(x) = \begin{cases} 
0 & \beta < 0\\
1 & \beta = 0\\
\infty & \beta > 0.
\end{cases}
$$
I suspect that you do not need to prove this statement formally: it is likely that there is a statement in the textbook that you can refer to.
With that established, address the problem in $3$ cases: in the case that $\alpha < 0$, conclude using the above fact that $\lim_{n \to \infty} z_n = \infty$, which means that the sequence is not bounded. In the case that $\alpha = 0$, conclude that $z_n \to 0$, which means that the sequence is convergent and is therefore bounded. Similarly, if $\alpha > 0$, conclude that $z_n \to 0$, which means that the sequence is convergent and therefore bounded.
Thus, we conclude that the sequence is bounded if and only if $\alpha \geq 0$.
