Limit of $(x_n)$ with $0
Let $0 < x_1 < 1$ and $x_{n + 1} = x_n - x_n^{n + 1}$ for $n \geqslant 1$.
Prove that the limit exists and find the limit in terms of $x_1$.
I have proved the existence but cannot manage the other part.
Thanks for any help.

Note that $x_2=x_1(1-x_1)$ with $0\lt x_1\lt1$ hence $0\lt x_2\lt1/4$, that $(x_n)_{n\geqslant1}$ is decreasing, in particular $(x_n)_{n\geqslant1}$ converges to some value $\ell(x_1)$ in $[0,x_1)$, and that $x_n\leqslant x_2$ for every $n\geqslant2$. Hence, for every $n\geqslant2$, $x_{n+1}\geqslant x_n-x_2^{n+1}$, which implies that $x_n\geqslant x_2-x_2^3-\cdots-x_2^n$ for every $n\geqslant2$ and that $\ell(x_1)\geqslant x_2-x_2^3/(1-x_2)=x_2(1-x_2-x_2^2)/(1-x_2)$ hence $\ell(x_1)\gt0$.
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To get an exact value of the limit $\ell(x_1)$ as an explicit function of $x_1$ is more difficult and, to venture a guess, probably not doable.

