<s>[INST] Show that a limit of a sequence is zero Let $(a_n)$ positive sequence where $$\dfrac{n-1}{n} \leq \dfrac{a_{n+1}}{a_n} \leq \dfrac {n}{n+1}$$
Prove that: $\lim_{n\to\infty} a_n = 0$.
I was stuck here:
I: $$\lim_{n\to\infty} \dfrac{n-1}{n} = \lim_{n\to\infty}1-\dfrac{1}{n} = 1$$
II: $$\lim_{n\to\infty} \dfrac {n}{n+1} = \lim_{n\to\infty} \dfrac{1}{1+ \dfrac{1}{n}} = 1$$
Then: $$\lim_{n\to\infty} \dfrac{a_{n+1}}{a_n} = 1$$
How do I continue from here?
 [/INST] Just observe that $$a_{n+1}\le \frac{n}{n+1}a_n\le \frac{n}{n+1}\cdot\frac{n-1}{n}a_{n-1}=\frac{n-1}{n+1}a_{n-1}\le\cdots\le \frac{1}{n}a_1\\a_{n+1}\ge \frac{n-1}{n}a_n\ge \frac{n-1}{n}\cdot\frac{n-2}{n-1}a_{n-1}=\frac{n-2}{n}a_{n-1}\ge\cdots\ge \frac{1}{n}a_2\\\implies \lim\sup_n a_n\le 0\le \lim\inf_n a_n\\\implies \lim_n a_n=0$$ Actually the lower bound on $a_n/a_{n-1}$ is redundant as $a_n$ is a positive sequence.
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