Convergence of $\sqrt[k]{z+\sqrt[k]{z+\sqrt[k]{z+\cdots}}}$, where $z=(1+x)^k-(1+x)$

Question

If one writes $$1+x=\sqrt{(1+x)^2}=\sqrt{1+2x+x^2}=\sqrt{x+x^2+(1+x)}$$ then one has a recursive definition of the function $1+x$ which can be used to write $1+x$ as the infinite nested radical: $$1+x=\sqrt{x+x^2+\sqrt{x+x^2+\sqrt{x+x^2+\sqrt{\cdot\cdot\cdot}}}}$$ But this definition relies on the fact that $$ 1+x=\sqrt{(1+x)^2} $$ which is only true for $x \ge-1$. In general one could state that $$ 1+x=\sqrt[n]{(1+x)^n}=\sqrt[n]{(1+x)^n-(1+x)+\sqrt[n]{(1+x)^n-(1+x)+\sqrt[n]{\cdots}}} $$ But the RHS does not converge to $1+x$ for most values of $x\in\mathbb{C}$. So, my question is, what is the actual closed form of the following function? For what values of $x\in\mathbb{C}$ does the following function converge? $$ \sqrt[k]{(1+x)^k-(1+x)+\sqrt[k]{(1+x)^k-(1+x)+\sqrt[k]{\cdots}}} $$ If unclear the above nested radical can be defined by $\lim_{n\to\infty}a_n$ where $$ a_1=\sqrt[k]{(1+x)^k-(1+x)},\quad a_n=\sqrt[k]{(1+x)^k-(1+x)+a_{n-1}}.$$

Answer 1

The expression $$ y=\sqrt[k]{(1+x)^k-(1+x)+\sqrt[k]{(1+x)^k-(1+x)+\cdots}} $$ where $x\ge 0$, represents the limit of the recursive sequence $$ a_1=\sqrt[k]{(1+x)^k-(1+x)}, \quad a_{n+1}=\sqrt[k]{(1+x)^k-(1+x)+a_n}, \quad n\in\mathbb N. $$ if such a limit exists.

Clearly, the sequence $\{a_n\}$ is increasing. (The fact $a_n\le a_{n+1}$ can be shown inductively.)

Next, we show inductively that $\{a_n\}$ is upper bounded by $1+x$. Clearly, $$ a_1=\sqrt[k]{(1+x)^k-(1+x)}\le\sqrt[k]{(1+x)^k}=1+x. $$ Assume that $a_n\le 1+x$. Then $$ a_{n+1}=\sqrt[k]{(1+x)^k-(1+x)+a_n}\le \sqrt[k]{(1+x)^k-(1+x)+(1+x)}=1+x. $$ Therefore, $\{a_n\}$ is increasing and upper bounded and thus it converges. Let $y=\lim a_n$.

If $x=0$, then observe that $a_n=0$, for all $n$, and hence $y=0$.

If $x>0$, then clearly $y>0$, and hence $$ y \leftarrow a_{n+1}=\sqrt[k]{(1+x)^k-(1+x)+a_n}\to \sqrt[k]{(1+x)^k-(1+x)+y} $$ and thus $$ y^k-y=(1+x)^k-(1+x) $$ Now the function $g(z)=z^k-z$ is strictly increasing, and hence one-to-one when $g'(z)=kz^{k-1}-1>0$ equivalently when $z>k^{-1/(k-1)}$. So if we show that $y>k^{-1/(k-1)}$, then we will shown that $y=1+x$.

We have $$ a_1=\sqrt[n]{(1+x)^k-(1+x)}=\sqrt[n]{(1+kx+\cdots)-(1+x)} \\ \ge \sqrt[n]{(k-1)x}\ge x^{1/k} $$ then $$ a_2=\sqrt[n]{(1+x)^k-(1+x)+a_1}\ge a^{1/k}_1\ge x^{1/{k^2}} $$ and in general $$ a_n\ge x^{1/{k^n}}\to 1, $$ and hence $$ y=\lim a_n\ge 1>k^{-1/(k-1)}, $$ which implies that $a_n\to 1+x$.

Note. If $x\in [-1,0]$, then $a_n\to 0$.