Limit of the sequence given by $x_{n+1}=x_n-x_n^{n+1}$

Question

Let , $x_1 \in (0,1)$ be a real number. For $n>1$ define $x_{n+1}=x_n-x_n^{n+1}$. Then prove that $\displaystyle \lim_{n \to \infty} x_n$ exists.

We have to prove that the given sequence $\{x_n\}$ is convergent. So we have to show that $\{x_n\}$ is monotone and bounded.

I proved that the sequence is monotone decreasing. But I'm unable to show that it is bounded below. How can I show it ?

Any other way to prove that the limit exists ?

Possible duplicate of Limit of $(x_n)$ with $0<x_1<1$ and $x_{n + 1} = x_n - x_n^{n + 1}$ — Arnaud D., Apr 18 '19 at 12:02

score 4 · Accepted Answer · answered Mar 17 '17 at 07:19

We show by induction that $x_n \in (0,1)$ for all $n$:

The case $n=1$ is clear.

Now let $n \in \mathbb N$ and $x_n \in (0,1)$

Then: $x_{n+1}=x_n(1-x_n^n)$. From $x_n \in (0,1)$ we get $x_n^n \in (0,1)$ and therefore $1-x_n^n \in (0,1)$.

Consequence: $x_{n+1} \in (0,1)$.

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