Find the limit of the sequence $x_{n+1}=x_n-x_n^{n+1}$ where $0<x_1<1$

Question

Find the limit of the sequence $x_{n+1}=x_n-x_n^{n+1}$ where $0<x_1<1$.

I think zero is the limit but how do I prove that? I prove that sequence is monotone decreasing and bounded so convergent.

Answer 1

Yes, the recursive sequence $\{x_n\}_{n\geq 1}$ is strictly decreasing and positive so it has a non negative limit $L(x_1)$. However $L(x_1)$ is always positive.

To see this, first note that $x_2=x_1-x_1^2\in (0,1/4]$. Moroever, for $n\geq 2$, $$x_n=x_2+\sum_{k=3}^n(x_k-x_{k-1})=x_2-\sum_{k=3}^nx_{k-1}^k\geq x_2-\sum_{k=3}^{\infty}x_{2}^k=x_2-\frac{x_2^3}{1-x_2}$$ Hence $$L(x_1)\geq x_2-\frac{x_2^3}{1-x_2}=\frac{x_2(1-(x_2+x_2^2))}{1-x_2}\geq \frac{11 x_2}{12}=\frac{11 x_1(1-x_1)}{12}>0.$$