I know by definition that a sequence converges if $\exists p \in X$ such that $\forall \epsilon > 0$, $\exists N \in \mathbb{R}$ s.t. $\forall n \in \mathbb{N}$ with $n>N$, $d(p_n, p) < \epsilon$. So I should find a $\epsilon$ that proves $| n (\sqrt{n^2+1} - n)-0| < \epsilon $.
Would it be correct to say that $\forall n>N$ such that $\sqrt{n^2+1} - n)$ < $\sqrt{N^2+1} - N) < \frac{1}{N} = \epsilon$? So in the end $N = \frac{1}{\epsilon}$ and this concludes that it does indeed converge to 0?
Given $\epsilon > 0$, we want to find $N$ such that
$$ \sqrt{n^2 + 1} - n < \epsilon$$ for all $n \geq N$. Note that $$ \sqrt{n^2 + 1} -n = \frac{1}{\sqrt{n^2+1} + n} < \frac1n.$$ Thus $$\sqrt{n^2 + 1} -n < \epsilon \quad \text{ for } \quad n\geq N = \frac1\epsilon.$$