Старая задач ч а с дополнением

[Rak so dna]

bot в сообщении #1719004 писал(а):

Тут у меня не кликается

Цитата:

The sequence was given as problem 4 at the final round of the 2024 Alibaba Mathematics Competition.
Here’s a solution to the following problem.

Цитата:

Let $a_1=c\in (0,1)$ and $a_{n+1}=a_n+\frac{a_n^2}{n^2}$. Prove that $\lim\limits_{n \to \infty} a_n=q<\frac{2}{1-c}$

It’s clearly true by induction that $a_k\leq kc$ for all $k$.
Now observe that

$$ \frac{1}{a_{n+1}}= \frac{1}{a_n}-\frac{n^2}{a_n+n^2} $$

$$ \frac{1}{a_{n+1}}-\frac{1}{n+1} = \frac{1}{a_n}-\frac{1}{n}+\frac{1}{n(n+1)}-\frac{n^2}{a_n+n^2}= \frac{1}{a_n}-\frac{1}{n}-\frac{n-a_n}{n(n+1)(a_n+n^2)} $$

Let $x_n= \dfrac{1}{a_n}-\dfrac{1}{n}$, this implies:

$$ x_{n+1} = x_n-\frac{x_n}{n(n+1)\left(\frac{1}{n}+\frac{n}{a_n}\right)}\geq x_n-\frac{x_n}{n(n+1)\left(\frac{1}{n}+1\right)}= x_n-\frac{x_n}{(n+1)^2} $$

So

$$ \frac{x_{n+1}}{x_n}\geq 1-\frac{1}{(n+1)^2},\ \forall n\geq1 $$

Consequently

$$ \frac{x_{n+1}}{x_1} \geq \prod_{k=1}^n \left(1-\frac{1}{(k+1)^2}\right)= \left( \frac{1 \cdot 3}{2^2} \right) \left( \frac{2 \cdot 4}{3^2} \right) \left( \frac{3 \cdot 5}{4^2} \right) \cdots \left( \frac{n(n+2)}{(n+1)^2} \right)=\frac{1(n+2)}{2(n+1)}\geq \frac{1}{2} $$

So

$$ \frac{1}{a_{n+1}}-\frac{1}{n+1}\geq \frac12\left(\frac{1}{a_1}-1\right)= \frac12\left(\frac{1-c}{c}\right)$$

Since $a_{n+1}\leq \left(\frac12 \left(\frac{1-c}{c}\right)+\frac{1}{n+1}\right)^{-1}$, the sequence $a_k$ is bounded above and increasing thus convergent, taking $n\to \infty$ gives $\lim\limits_{n\to \infty}a_n\leq \frac{2c}{1-c}<\frac{2}{1-c}$. This completes the proof.

(Remark)

Personally, I think it’s unlikely that the constant in the problem statement can be improved to $\frac32$.

Но тут вроде как ошибка во второй строке:

$$ \frac{1}{a_{n+1}}-\frac{1}{n+1} = \frac{1}{a_n}-\frac{1}{n}+\frac{1}{n(n+1)}-\frac{n^2}{a_n+n^2}= \frac{1}{a_n}-\frac{1}{n}-\frac{n-a_n}{n(n+1)(a_n+n^2)}\color{red}-\frac{n^2-1}{a_n+n^2} $$