Estimating the limit $x_{n+1} =x_n - x_{n}^{n+1} $

Question

I wonder whether there is a general method for accurately estimating the limit of the sequence:

\begin{equation} x_{n+1} = x_n - x_{n}^{n+1}, \forall x_1 \in (0,1) \end{equation}

After showing that the limit exists, since $ x_n $ is decreasing and bounded, I managed to derive a lower-bound. In particular, I used the fact that:

\begin{equation} \frac{x_{n+1}}{x_n} = 1-x_{n}^n \tag{1} \end{equation}

Using $(1)$ we obtain:

\begin{equation} \frac{x_N}{x_{N-1}}...\frac{x_2}{x_1}=\prod_{n=1}^{N} (1-x_{n}^n)=\frac{x_N}{x_1} \tag{2} \end{equation}

From this we deduce:

\begin{equation} \begin{split} \lim_{N \to \infty} x_N & = \lim_{N \to \infty}x_1 \prod_{n=1}^{N} (1-x_{n}^n) \\ & = x_1 (\lim_{N \to \infty} \prod_{n=1}^{N} e^{\ln (1-x_{n}^n)}) \\ & = x_1 (\lim_{N \to \infty} e^{\sum_{n=1}^N\ln (1-x_{n}^n)}) \end{split} \tag{3}\end{equation}

Using the following facts:

\begin{cases} \sum_{n=1}^{N} \ln(1-x_{n}^n) \geq \sum_{n=1}^{N} \ln(1-x_{1}^n),\\ x \approx 0 \implies \ln(1+x) \approx x \\ \tag{4}\end{cases}

We may deduce that for $M$ sufficiently large:

\begin{equation} \sum_{n=1}^{\infty}\ln (1-x_{n}^n) \geq \sum_{n=1}^{M} \ln(1-x_{1}^n)-\sum_{n=M}^\infty x_{1}^n \tag{5} \end{equation}

And using $(5)$ we have a useful lower-bound. However, I wonder whether there's a more direct integration technique which can give me a good approximation to $(3)$.

Answer 1

The observation of Fede Poncio that the limit of $x_n$ is well-approximated by a quadratic polynomial in terms of $x_1$ is very useful. With a few lines of Python3 code we observe:

import numpy as np
import matplotlib.pyplot as plt

def f(x,N):
    X = np.zeros(N,dtype='float64')

    X[0] = x

    for i in range(1,N):
        X[i] = X[i-1] - X[i-1]**(i+1)

    return X, X[-1]

z= np.linspace(0,1,100)

q = [f(z[i],1000)[1] for i in range(100)]

plt.xlabel('unit interval')

plt.ylabel('approximate limit')

plt.plot(z,q,color='steelblue')

Indeed, this observation may be used to derive a very good approximation.

By expanding $x_{N+1}$ we obtain:

\begin{equation} \begin{split} x_{N+1} & = x_1-x_1^2-x_2^3-x_3^4-...-x_N^{N+1} \\ & = (x_1-x_1^2)-\sum_{n=2}^N x_n^{n+1} < x_1-x_1^2=x_2 \\ \end{split} \tag{1} \end{equation}

Now, given that $x_n$ is decreasing:

\begin{equation} \sum_{n=2}^{N} x_n^{n+1} < x_2 \sum_{n=2}^{\infty} x_2^n=x_2\big(\frac{x_2}{1-x_2}-x_2 \big)=\frac{x_2^3}{1-x_2}<2x_2^3 \tag{2} \end{equation}

In fact, we may show that:

\begin{equation} \lim_{n\to\infty} x_n \sim x_1-x_1^2 \tag{3} \end{equation}

The quality of this approximation can be checked using $(1)$ and $(2)$ as follows:

\begin{equation} \big\lVert 1-\frac{\lim_{n\to\infty} x_n}{x_1-x_1^2} \big\rVert \leq \frac{2x_2^3}{x_2}=2x_2^2 \leq \frac{2}{4^2}=12.5 \% \tag{4} \end{equation}

Moreover, if we calculate the expected value of $(4)$ we find:

\begin{equation} \mathbb{E}\big[\big\lVert 1-\frac{\lim_{n\to\infty} x_n}{x_1-x_1^2} \big\rVert\big] \leq \frac{1}{15} \sim 6 \% \tag{5} \end{equation}

Answer 2

As you noted, $(x_n) $ is convergent as a decreasing positive sequence.

We should have $$\lim_{n\to\infty}x_n^{n+1}=$$ $$\lim_{n\to\infty}e^{(n+1)\ln (x_n)}=0$$

for $A <0$ ans great enough $n,$

$$\ln (x_n)<\frac {A}{n+1}$$