# Proof for the Ratio Test using the squeeze theorem

Excerpt from my University's lecture notes

The assumption in the excerpt above is that $$a_n >0$$.

In the above image, at the bottom it mentions "so $$a_n \rightarrow 0$$ by the sandwich theorem" but I don't see anywhere that I could apply the squeeze theorem. Is the squeeze theorem being applied to this set of inequality? $$0\leq a_{N+s} \leq y^{N+s}\left(\frac{a_N}{y^N}\right)$$

If I'm missing out any piece of information, please let me know. I've been trying to understand this for hours and I've looked at many resources but I do not understand where or how the squeeze theorem is being used. Thank you in advance.

Yes, the squeeze theorem implies that $$\lim_{s \to \infty} a_{N+s} = 0$$ since the we can 'squeeze it' between two sequences which converge to zero as $$s \to \infty.$$ These are the constant sequence $$0$$ on the left, and $$y^sa_N =y^{N+s}\left(\frac{a_N}{y^N}\right)$$ on the right.
From $$\lim_{s \to \infty} a_{N+s} = 0$$ it easily follows that $$\lim_{n \to \infty} a_n=0$$, as you can check from the definition of the limit of a sequence.
• If the ratio test $lim_{n \to +\infty} \frac{a_{n+1}}{a_n} = L > 1$, we can show that for some $C > 0, \infty > a_n \geq Cy^n$ so we can apply the squeeze theorem to show that $a_n$ tends to infinity. Is my reasoning solid here? Jan 20, 2021 at 1:53
• @Glacey yes, if you have proved a squeeze theorem for infinity and that $\lim_{n \to \infty} y^n = \infty$ if $y>1.$ A 'squeeze theorem for infinity' is very easy to prove, you could try to prove it as an exercise. Jan 20, 2021 at 8:04