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Let $\{a_n\}_{n\in\mathbb{N}}$ be a sequence of real numbers such that:

  1. $a_n\in[0,1]$, $\forall n\in\mathbb{N}$
  2. $\lim_{n\to\infty}a_n = 0$
  3. $\lim_{n\to\infty}\frac{a_{n+1}}{a_n} = 1$
  4. $a_{n+1} \leq a_n$, $\forall n\in\mathbb{N}$ (decreasing sequence)
  5. $\frac{a_{n+1}}{a_n} \geq \frac{a_{n}}{a_{n-1}}$ (increasing sequence)

Prove or disprove that: $$ b_n = na_n \mbox{ is eventually monotone} $$ so that there exists $n_0\in\mathbb{N}$ such that $\{b_n\}_{n\geq n_0}$ is a monotone sequence.

Context

Unfortunately I haven't an "analalytic form" for the sequence $a_n$... it can not be computed explicitely. Although I know that it converges to 1 and that it is bounded in $[0,1]$, and that the quotient between consecutive terms goes to $1$ (with an increasing behavior).

Idea

One idea to prove that $\{b_n\}$ is eventually monotone, is to prove that: $$ \frac{b_{n+1}}{b_n} \geq 1 \mbox{ or } \frac{b_{n+1}}{b_n} \leq 1 \mbox{ $\forall n\geq n_0$} $$ so that: $$ \frac{(n+1)a_{n+1}}{na_n} \geq 1 \mbox{ or } \frac{(n+1)a_{n+1}}{na_n} \leq 1 \mbox{ $\forall n\geq n_0$} $$

The main problem is that I don't know the rate of convergence of $\frac{a_{n+1}}{a_n}$ to $1$, I only know that $\frac{a_{n+1}}{a_n}$ is monotone increasing.

MathRevenge
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    Just a disclaimer: I haven't downvoted to give a chance to improve the question, but as a longer term user: You will probably already know by now that you have to show your efforts/ give context. – J.D May 20 '24 at 18:15
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    Sorry, I've just returned home.. I'll write it better. Thank you for pointing it out. – MathRevenge May 20 '24 at 18:23

1 Answers1

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The statement is false, $b_n$ can be non monotone. I give you a counterexample.

Consider the sequence $a_n$ recursively defined as

$$ a_0 = 1,$$

$$a_{2m+1} = a_{2m}\cdot \frac{12m^2+12m+1}{12m^2+18m+6}, \quad\text{for $m \geq 0$},$$

and

$$ a_{2m}=a_{2m-1}\cdot\frac{12m^2-12m+2}{12m^2-6m}, \quad\text{for $m \geq 1$}.$$

We firstly prove that $\{a_n\}$ satisifes the hypothesis, and then that $b_n$ is not monotone. First define

$$ \alpha_m = \frac{12m^2+12m+1}{12m^2+18m+6}$$

and

$$ \beta_m = \frac{12m^2-12m+2}{12m^2-6m},$$

so that $a_{2m+1} = \alpha_m \cdot a_{2m}$ and $a_{2m} = \beta_m \cdot a_{2m-1}$. You can easily check that $0 < \alpha_m< 1$ for any $m \geq 0$, and $0 < \beta_m< 1$ for any $m \geq 1$, so you have (1) for free. To prove that (2) is satisifed by $\{a_n\}$, notice that for $m > 1$

$$ a_{2m} \leq \beta_1 \dots \beta_m \qquad \text{and}\qquad a_{2m+1} \leq \alpha_0 \dots \alpha_m.$$

Hence it's sufficient to prove that the two products go to zero as $m \rightarrow \infty$. We can write

$$ \alpha_k = 1 - \frac{6k + 5}{12k^2 + 18k + 6}$$

and

$$ \beta_k = 1 - \frac{6k - 2}{12k^2 -6k}.$$

By this, it's then sufficient to prove that

$$ \sum_{k = 1}^{\infty} \frac{6k + 5}{12k^2 + 18k + 6} $$

and

$$ \sum_{k = 1}^{\infty} \frac{6k - 2}{12k^2 -6k} $$

both diverge, which is seen to be true by standard arguments using the harmonic series. We can conclude that $a_n \rightarrow 0$ as $n\rightarrow \infty$ as the sequence is dominated by a sequence going to zero. Hence (2) holds.

Condition (3) is much easier, since $\frac{a_{2m+1}}{a_{2m}} = \alpha_m$, $\frac{a_{2m}}{a_{2m-1}} = \beta_m$, and $\alpha_m, \beta_m \rightarrow 1$.

Condition (4) holds again because $\alpha_m, \beta_m < 1$.

Condition (5) is

$$ \alpha_m \geq \beta_m \qquad \text{and}\qquad \beta_m \geq \alpha_{m-1}.$$

The first inequality is

$$ \frac{12m^2+12m+1}{12m^2+18m+6} \geq \frac{12m^2-12m+2}{12m^2-6m}, $$

while the second is

$$ \frac{12m^2-12m+2}{12m^2-6m} \geq \frac{12(m-1)^2+12(m-1)+1}{12(m-1)^2+18(m-1)+6}. $$

You can check that they both hold for $m \geq 1$ by hand (or choose your favorite solver if you are as lazy as me).

So we have a "right" sequence. Moreover, we have

$$ \frac{b_{2m+1}}{b_{2m}} = \frac{2m+1}{2m}\alpha_m = \frac{(2m+1)(12m^2+12m+1)}{2m(12m^2+18m+6)} > 1$$

and

$$ \frac{b_{2m}}{b_{2m - 1}} = \frac{2m}{2m-1}\beta_m = \frac{2m(12m^2-12m+2)}{(2m-1)(12m^2-6m)} < 1$$

again using our never-loved-enough solver.

I hope I did't make any mistake (WARNING: HIGH POSSIBILITY, PLEASE CHECK IT).

However, you might wonder how did I come to such a sequence. The trick is splitting it into two sequences depending on $n$ being even or odd, assuming $a_{2m+1} = \alpha_m a_{2m}$ and $a_{2m} = \beta_m a_{2m-1}$ for some $\alpha_m$ and $\beta_m$ and then carefully write all your conditions in this way.

2by2is2mod2
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