Proving a sequence to be divergent

Question

I'm trying to prove this sequence: $a_n = \sqrt{n}-\sqrt{n^2-1}$ to be divergent. How would I do this? I'm thinking of proving that it's not bounded below, but I'm not sure how to do that with induction, as I've only done that to prove it's bounded.

Answer 1

$$ a_n = \sqrt{n} - \sqrt{n^2-1} = \sqrt{n} - n\sqrt{1-\frac{1}{n^2}} = \sqrt{n}\underbrace{\left( 1-\sqrt{n}\sqrt{1-\frac{1}{n^2}}\right)}_{b_n} $$ Now, $\sqrt{n}\xrightarrow[n\to\infty]{}\infty$; and $\sqrt{1-\frac{1}{n^2}}\xrightarrow[n\to\infty]{}1$, so $\sqrt{n}\sqrt{1-\frac{1}{n^2}}\xrightarrow[n\to\infty]{}+\infty$ and thus $b_n\xrightarrow[n\to\infty]{}-\infty$.

Finally, this implies that $a_n=\sqrt{n}b_n\xrightarrow[n\to\infty]{}-\infty$.

Answer 2

Hint: compare the sequence $a_n=\sqrt{n}-\sqrt{n^2-1}$ to the sequence $b_n=-\sqrt{n}$, which clearly diverges to $-\infty$.