# Proof verification for $\lim_{n\to\infty}(\sqrt{n^2-1} - \sqrt n) = +\infty$

Show that: $$\lim_{n\to\infty}\left(\sqrt{n^2-1} - \sqrt n\right) = +\infty$$

I've started it this way.

Lemma:

Let $$x_n$$ and $$y_n$$ be two sequences. Claim:

If: $$\begin{cases} &\lim_{n\to\infty} x_n =+\infty \\ &\exists N\in \Bbb N, \ \forall n >N:y_n\ge c > 0 \end{cases}$$ Then: $$\lim_{n\to\infty}(x_ny_n) = +\infty$$

Proof:

$$\Box$$ Start with definition of limit for this case: $$\forall\varepsilon>0,\ \exists N_1\in\Bbb N: \forall n > N_1 \implies x_n >\varepsilon$$ Also: $$\exists N_2\in\Bbb N:\forall n>N_2 \implies y_n \ge c > 0$$

Let: $$N = \max\{N_1, N_2\}$$

Then starting from this $$N$$ we obtain: $$x_n\cdot y_n > c\cdot \varepsilon$$

And we have that: $$\forall\varepsilon>0,\ \exists N =\max\{N_1, N_2\}\in\Bbb N: \forall n > N \implies x_n y_n > c\varepsilon$$

Thus: $$\lim_{n\to\infty}(x_ny_n) = +\infty \ \Box$$

Now back to the initial problem. Let: $$z_n = \sqrt{n^2-1} - \sqrt n = \frac{n^2 - n - 1}{\sqrt{n^2 - 1} + \sqrt{n}}$$

Define: $$x_n = n - 1 - {1\over n} \\ y_n = \frac{n}{\sqrt{n^2 - 1} + \sqrt{n}}$$

Obviously $$y_n \ge c > 0$$ for some $$N$$ and $$n>N$$. Also $$x_n \to +\infty$$, then by lemma: $$\lim_{n\to\infty}z_n = \lim_{n\to\infty}{x_ny_n} = +\infty$$

I know this is a bit overkill, but i wanted to use that exact lemma for the proof. Apart from that, is it valid?

BTW here is a visualization for $$x_n, y_n$$

Update

Since it is not clear where the lemma comes from here is the problem from the problem book right before the limit.

Let: $$\lim_{n\to\infty}x_n = a\ , \text{where}\ a = +\infty \ \text{or} \ a = -\infty$$ Prove that if for all $$n$$ starting from some $$N$$ $$y_n \ge c > 0$$ then $$\lim_{n\to\infty}x_ny_n = a$$ And if for all $$n$$ starting from some $$N$$ $$y_n \le c < 0$$ then $$\lim_{n\to\infty}x_ny_n = -a$$

No other constraints are given.

• What's $c$ in your lemma? It feels like it's just randomly introduced? – Jam Dec 6 '18 at 19:30
• @Jam, the exercise to prove this lemma comes right before the exercise on the limit (which in in the question section). No constraints for $c$ are given except for the fact that from some $N$ $y_n \ge c > 0$. This lemma is actually one of the several of the same kind in the exercise before limit. – roman Dec 6 '18 at 19:32
• I may be mistaken but I think you need more constraints on $c$. It seems like there's nothing stopping us from having $c=1/\varepsilon>0$, in which case $x_ny_n>c\varepsilon=1$ doesn't tell us much. – Jam Dec 6 '18 at 19:39
• In other words, you've shown that $x$ is unbounded above and $y$ is bounded from below but how do we know that all $y$ have the same lower bound? And how do we know that the lower bound of $y$ doesn't get small very quickly (counteracting the size of $\varepsilon$)? – Jam Dec 6 '18 at 19:42
• @Jam, I have updated the question and added more context of where that lemma came from – roman Dec 6 '18 at 22:10

1. $$\sqrt{n^2-1}\ge n-1$$ if $$n\ge 1$$ (to see this, just square both sides);
2. $$\sqrt n\le n/2$$ if $$n\ge 4$$ (to see this, just square both sides).
So $$\sqrt{n^2-1}-\sqrt n \ge n/2-1$$ if $$n\ge 4$$.
And $$\lim_{n\rightarrow\infty}(n/2-1) = +\infty$$.