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?


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.

  • $\begingroup$ Can you explain why $\lim \sup a_n \leq 0 \leq \lim \inf a_n$ ? $\endgroup$ – Dan Revah Jan 10 '17 at 19:17
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    $\begingroup$ From the first set of inequalities you have $\lim\sup_n a_{n+1}\le \lim_{n\to \infty}a_1/n=0$, and similar reasoning for the lim inf case. $\endgroup$ – Samrat Mukhopadhyay Jan 10 '17 at 19:18

@Samrat Mukhopadhyay Another solution, in the same vein as yours:

Let us define: $b_n:=na_n$.

The second inequality can thus be written: $b_{n+1}\leq b_n.$

Thus $b_n$ is a decreasing sequence of positive numbers.

Being bounded from below by zero, sequence $b_n$ converges to its "lim inf", say $M \geq 0$.

From $\lim_{n \to \infty} b_n = M$ we conclude that, using the definition of $b_n$,

$$\lim_{n \to \infty} a_n = \lim_{n \to \infty} b_n \dfrac{1}{n} = M\times 0 = 0.$$

Remark: surprisingly, the first inequality doesn't play any role.

  • $\begingroup$ The first inequality would've implied the assumption that the sequence is positive. But you given the sequence is assumed positive, it oughtn't be that shocking that a lower bound proved unnecessary for establishing the limit as $0$. $\endgroup$ – AJY Jan 10 '17 at 20:31
  • $\begingroup$ @AJY, actually the first inequality only implies the terms all have the same sign. The proof here could be made to work with $b_n:=n|a_n|$. $\endgroup$ – Barry Cipra Jan 10 '17 at 20:52
  • $\begingroup$ Right you are. But it would imply that the sequence still goes to $0$ even if we don't assume nonnegative. $\endgroup$ – AJY Jan 10 '17 at 20:53

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