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.


1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$
    – Glacey
    Jan 20, 2021 at 1:53
  • $\begingroup$ @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. $\endgroup$
    – Steven
    Jan 20, 2021 at 8:04

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