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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?

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    $\begingroup$ In order for us to tell you how to write things out, it would be helpful if you explained why you believe that the answer is what it is. Why do you think that $\frac 1{n^{\alpha}}$ converges for $\alpha \geq 0$? Are you saying that $\frac 1{n^{\alpha}}$ is not bounded when $\alpha < 0$? If so, then you must say so explicitly in your answer. Also, why do you believe that this is the case? $\endgroup$ – Ben Grossmann Aug 16 '20 at 10:16
  • $\begingroup$ Because it is clear that for $\alpha\geq0$ the sequence converges to 0. If $\alpha<0$ then the value of $\frac{1}{n^\alpha}$ will become very big unless $\alpha>-\frac{1}{n}$. I might be wrong, but this is what I think. I don't know how to approach this question. $\endgroup$ – Jess Aug 16 '20 at 10:20
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    $\begingroup$ By giving the answer that you have given, you have not only "approached" the problem correctly, but also have given an almost complete answer. It seems that your only question, then, is how to write this up with sufficient "formality." $\endgroup$ – Ben Grossmann Aug 16 '20 at 10:25
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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.

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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 $\alpha \geq 0$.

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  • $\begingroup$ If the sequence is bounded, does that mean that the limit of the sequence exists? Because I just noticed that the next question asks me to find the values of $\alpha$ for which lim$_{n\to\infty}z_n$ exists. I assume the values of $\alpha$ are the same as which $z_n$ is a bounded sequence? $\endgroup$ – Jess Aug 16 '20 at 10:40
  • $\begingroup$ It is not necessarily the case that a bounded sequence will have a limit. However, we can clearly see by exhausting the possibilities that for this problem, the sequence $z_n$ converges whenever it is bounded. $\endgroup$ – Ben Grossmann Aug 16 '20 at 10:42

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