Null Sequences and Real Analysis I came across the following problem during the course of my study of real analysis:

Prove that $(x_n)$ is a null sequence iff $(x_{n}^{2})$ is null.

For all $\epsilon>0$,  $|x_{n}| \leq \epsilon$ for $n > N_1$. Let $N_2 = \text{ceiling}(\sqrt{N_1})$. Then $(x_{n}^{2}) \leq \epsilon$ for $n > N_2$. If $(x_{n}^{2})$ is null then $|x_{n}^{2}| \leq \epsilon$ for $n>N$. Let $N_3 = N^2$. Then $|x_n| \leq \epsilon$ for $n> N_3$. 
Is this correct? In general, we could say $(x_{n})$ is null iff $(x_{n}^{n})$ is null?
 A: That's not quite correct since you don't know the connection between $n$ and $x_n$ - hence taking square roots of $n$ are meaningless. What you can do is the following: 


*

*$x_n\to 0$, hence for all $\epsilon>0$ exists $N$ such that $|x_n|<\sqrt{\epsilon}$ if $n\geq N$, but then it means that  for all $\epsilon>0$ exists $N$ such that $|x_n|^2<\epsilon$ if $n\geq N$.

*$x_n^2\to 0$, hence for all $\epsilon>0$ exists $N$ such that $|x_n|^2<\epsilon^2$ if $n\geq N$, but then it means that  for all $\epsilon>0$ exists $N$ such that $|x_n|<\epsilon$ if $n\geq N$.
A: Concerning the last question, it is true that, for any fixed positive integer $k$, $x_n \to 0$ if and only if $x_n^k \to 0$. Also, $x_n \to 0$ obviously implies that $x_n^n \to 0$. On the other hand, $x_n^n \to 0$ does not imply $x_n \to 0$ (take for example $x_n = 1/2$ $\forall n$).
A: You could use the following fact:
If a function $f:X\to Y$ between two topological spaces is continuous and $x_n\to x$, then $f(x_n)\to f(x)$.
(In case you do not have learned it in this generality, you might at least know that this is true for real functions or for functions between metric spaces. In fact, in the case of real functions the above condition is equivalent to continuity.)
You can obtain your first claim by applying the fact to the continuous functions:
$f: \mathbb R\to\mathbb R$, $f(x)=x^2$ (one implication)
$f: \langle 0,\infty)\to \mathbb R$, $f(x)=\sqrt{x}$ (reverse implication)
