# Show that $\lim_{n\to\infty} (\sqrt{2n} - \sqrt{n - 1}) = \infty$

(Limit of succession) Prove that $$\lim_{n\to\infty} (\sqrt{2n} - \sqrt{n - 1}) = \infty$$

My idea is: $$\lim_{n\to\infty} (\sqrt{2n} - \sqrt{n - 1}) = \lim_{n\to\infty} (\sqrt{2n} - \sqrt{n - 1}) \frac{(\sqrt{2n} + \sqrt{n - 1})}{ \sqrt{2n} + \sqrt{n - 1}} = \lim_{n\to\infty} \frac{n + 1}{\sqrt{2n} + \sqrt{n - 1}}$$

But I can't get the final steps...I tried using the limit comparison test but without success.

EDIT

We didn't show that $$\lim_{n\to\infty} \sqrt{a_n} = \sqrt{\lim_{n\to\infty} a_n}$$ so I have to show it (how? Maybe using the definition of limit?) or I should consider to solve my problem in another way (best way seems notice that $\sqrt {2n} - \sqrt{n-1} > \sqrt{2n} - \sqrt{n}$ as @астон вілла олоф мэллбэрг suggested)

• how about dividing top and bottom by root n - from where you got to – Cato Nov 17 '16 at 12:48
• divide by $\sqrt{n}$ – AspiringMat Nov 17 '16 at 12:52

Simply note that $\sqrt {2n} - \sqrt{n-1} > \sqrt{2n} - \sqrt{n}$, and $\sqrt{2n} - \sqrt{n} = (\sqrt 2 - 1)(\sqrt n)$, which is a constant times $\sqrt n$, so goes to infinity as $n$ goes to infinity.
You are on the right track. Note that $$\frac{n + 1}{\sqrt{2n} + \sqrt{n - 1}}=\frac{n}{\sqrt{n}}\cdot\frac{1 + \frac{1}{n}}{\sqrt{2} + \sqrt{1 - \frac{1}{n}}}.$$ The term $\frac{n}{\sqrt{n}}=\sqrt{n}$ goes to infinity and the other one goes to $1/(\sqrt{2}+1)>0$.
While the answer by астон вілла олоф мэллбэрг is the best way to solve this limit, here's a way to continue from your last step: $$\sqrt{2n} + \sqrt{n-1} \leq \sqrt{2n+2} + \sqrt{n+1} = (\sqrt{2}+1)\sqrt{n+1}$$ Therefore $$\frac{n+1}{\sqrt{2n} + \sqrt{n-1}} \geq \frac{n+1}{(\sqrt{2}+1)\sqrt{n+1}} = \frac{\sqrt{n+1}}{(\sqrt{2}+1)}$$ Then $$\infty = \lim_{n \to \infty} \frac{\sqrt{n+1}}{(\sqrt{2}+1)} \leq \lim_{n \to \infty} \frac{n+1}{\sqrt{2n} + \sqrt{n-1}} \leq \infty$$
Hence your limit must be $\infty$ too.