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A null sequence in $\mathbb{R}$ is a sequence $(a_n)_{n\in\mathbb{N}}$ , so that: $\forall\varepsilon>0, \exists n_0\in\mathbb{N},\forall n\in\mathbb{N}_{\geq n_0} : \left | a_n \right |< \varepsilon $

$a$ is called a limit of $(a_n)$ if $(a_n-a)$ is a null sequence.

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    Yes, it's fine. You can show it's equivalent to the usual definition. – 2017-02-10
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    Put differntly: What makes you think that it is not ok? – 2017-02-10

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The first part is okay. Since we say that the sequence $(x_n)_n$ converges to $x$ if and only if for all $\varepsilon > 0$ there exists a number $N(\varepsilon) \in \mathbf N$ such that $|x_n -x| <\varepsilon$ for all $n \geq N(\varepsilon)$. In this case we write $\lim_n x_n =x$ or $x_n \to x$. In your special case set $x=0$. Such a sequence is called null sequence.

The second part is fine aswell since we have $$x_n \to x \qquad \Longleftrightarrow \qquad (x_n-x)_n \ \text{is a null sequence}.$$

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Some formality:

Def: 1) A null sequence in $\mathbb{R}$ is a sequence $(a_n)_{n\in\mathbb{N}}$ , so that: $\forall\varepsilon>0, \exists n_0\in\mathbb{N},\forall n\in\mathbb{N}_{\geq n_0} : \left | a_n \right |< \varepsilon $

2) Given $(b_n)_{n\in\mathbb{N}}$ a real sequence, and $b\in \mathbb{R}$ s.t., if we define $a_n:=(b_n-b)$ and then we have that $(a_n)_{n\in\mathbb{N}}$ is a null sequence, we'll say that $b$ is the limit of $(b_n)_{n\in\mathbb{N}}$