2

$x_{n+1}=\frac{p-1}{p} x_{n}+\frac{\alpha}{p} x_{n}^{-p+1},p\in N^+\ge2, \alpha>0,x_{1}>\alpha^\frac{1}{p},prove \lim_{n\to\infty}{x_n}=\alpha^\frac{1}{p} $

how to solve it ? it's too difficult to me.

after get this inequation , I have no clue to do next step.

$\frac{1}{P}\left((p-1) x_{n}+a x_{n}^{-p+1}\right)>\sqrt[p]{(p-1) a x_{n}^{-p+2}}$

nevermind_15
  • 589

2 Answers2

1

I think there is a typo: it seems that every $\alpha$ should be changed to $a$.

If we show that the limit exists and is $\alpha$, then the claim follows from $$ \alpha= \frac{p-1}{p}\alpha + \frac{a}{p} \alpha^{-p+1}.$$

We show that indeed $\{x_n\}$ converges.

  1. By AM-GM inequality, $$ x_{n+1}= \frac{(x_n+\cdots x_n) + ax_n^{-p+1}}{p} \geq \sqrt[p]{a}.$$ Since $x_1>\sqrt[p]{a}$ by assumption, it follows that $x_n\geq \sqrt[p]{a}$ for all $n \in \mathbb N$.

  2. Note that $$ x_n-x_{n+1}=\frac{x_n-a x_n^{-p+1}}{p}=\frac{x_n}{p} (1-ax_n^{-p}) \geq 0$$ since $x_n\geq \sqrt[p]{a}$.

From the monotone convergence claim $\{x_n\}$ converges.

Laplacian
  • 2,494
1

Finding the limit (assuming it exists): To facilitate the analysis, define the function:

$$F(x) \equiv 1 - \frac{1}{p} \bigg( 1 - \frac{a}{x^{p}} \bigg) \quad \quad \quad \text{for all } x > 0.$$

Since the first term in your sequence is positive, the recursive equation guarantees that all the terms are positive. Thus, rearranging your recursive expression gives the equivalent form:

$$F(x_{n}) = \frac{x_{n+1}}{x_n}.$$

Now, if the limit $\lim_{n \rightarrow \infty} x_n = L$ exists, then taking the limit of both sides of this equation (and noting that $F$ is a continuous function) gives $F(L) = 1$, which rearranges to give the desired solution $L = a^{1/p}$. This establishes the desired limit, assuming it exists. Thus, it remains only to show that the limit exists (i.e., the sequence converges). This can be accomplished using the monotone convergence theorem, by showing that the values in the sequence are strictly decreasing down to the limit. Proof of these two elements is below.

Lower bound: If $x > L$ then it is easy to establish that $x F(x) > L$. This implies that if a value in the sequence is above the limit value, the next value in the sequence (and thus all subsequent values) will also be above the limit value. Since $x_1 > L$ this establishes that $x_n > L$ for all $n \in \mathbb{N}$.

Monotonicity: Since $p>0$ we have $F'(x) = - a/x^{p+1} < 0$, which means that the function $F$ is strictly decreasing. Since $F(L) = 1$ this means that $F(x) < 1$ for all $x > L$, which means that if a value in the sequence is above the limit, the next value in the sequence will be less than that value. We have already established that all values in the sequence are above the limit, so this is a monotone decreasing sequence.

Ben
  • 4,079