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So we were doing a question at university, where, we had to show the sequences tended to $0$ using the squeeze theorem or otherwise. $$a_n = \begin{cases} \frac{1}{n} & n = 2m + 1 \\ \frac{-1}{n} & n = 2m \end{cases}$$ The way we tried applying the squeeze theorem is to write the function as $\frac{(-1)^n}{n}$ and then to apply the squeeze theorem say, $$-1 \leq (-1)^n \leq 1$$ Dividing throughout by $n$, we get, $$\frac{-1}{n} \leq \frac{(-1)^n}{n} \leq \frac 1 n$$ Now, clearly applying limits throughout, $$\lim_{n \rightarrow \infty}\frac{-1}{n} \leq \lim_{n \rightarrow \infty}\frac{(-1)^n}{n} \leq \lim_{n \rightarrow \infty}\frac 1 n$$ Which gives us, $$\lim_{n \rightarrow \infty} a_n = 0$$ Is this an acceptable proof, someone pointed out that you couldn't write, the first statement since the function $(-1)^n$ only has two values $\{-1,1\}$ and that the middle function touches both at infinitely many points so you couldn't apply the squeeze theorem. Any input would be appreciated.

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  • $\begingroup$ You mean $a_n=\dfrac{(-1)^{n+1}}{n}$ ? $\endgroup$ – Aryadeva Oct 16 '20 at 21:10
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    $\begingroup$ @Aryadeva Comment withdrawn, you are right. To the OP: In my opinion, your proof is perfectly valid, except for the typo that Aryadeva's comment focuses on. $\endgroup$ – user2661923 Oct 16 '20 at 21:13
  • $\begingroup$ No problem @user2661923 $\endgroup$ – Aryadeva Oct 16 '20 at 21:17
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    $\begingroup$ @Aryadeva yeah my bad, that's a typo it's supposed to be the other way aroudn $\endgroup$ – Prakhar Nagpal Oct 16 '20 at 21:21
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    $\begingroup$ "touches both at infinitely many points so you couldn't apply the squeeze theorem" - this simply is not true, your proof is fine. $\endgroup$ – Sil Oct 16 '20 at 21:54
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$$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}=0$$ By definition means that for any $\epsilon>0$ there exists an $N$ such that if $n>N$ then $$\left|\frac{1}{n}\right|<\epsilon\quad\quad (1)$$

Thus obviously if $a_n=\frac{(-1)^{n+1}}{n}$ then $a_n\to 0$ as $n\to\infty$ because of the absolute value in the definition $(1)$.

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That guy didn't consider that you apply the squeeze theorem only in the last two lines. Other inequalities are perfectly innocent.

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