$x_{n+1}=-1+\sqrt[n]{1+nx_n}$, $x_1>0$ limits

In this answer, we will prove:

$$ \lim_{n\to\infty} n x_n = 0 \qquad\text{and}\qquad \lim_{n\to\infty} \frac{x_{n+1}}{x_n} = 1. \tag{*} $$

In fact, we will establish a stronger statement:

$$ \lim_{n\to\infty} n^2 x_n = 4. $$

Note that this implies $\text{(*)}$.

Lemma 1. $(x_n)$ is monotone decreasing and

$$ \bbox[border:1px green solid;padding:8px;background-color:#e8ffe8;]{\lim_{n\to\infty} x_n = 0.} $$

Proof. By the Bernoulli's inequality, we get

$$ x_{n+1} \leq \left(1 + \frac{1}{n} \cdot nx_n\right) - 1 = x_n $$

and hence $(x_n)$ is monotone decreasing. Then, from $x_n \leq x_1$,

$$ 0 \leq x_{n+1} \leq \sqrt[n]{1 + n x_1} - 1. $$

By letting $n \to \infty$ and applying the squeezing lemma, we conclude $\lim x_n = 0$. $\square$

Lemma 2. Define $\psi : [0, \infty) \to [1, \infty)$ by $\psi(x) = \int_{0}^{1} (1 + x)^{t} \, \mathrm{d}t = \frac{x}{\log(1+x)}$. Then

$$ \bbox[border:1px green solid;padding:8px;background-color:#e8ffe8;]{\frac{x_n}{x_{n+1}} \leq \psi(nx_n) \leq \frac{x_n}{x_{n+1}}(1+x_{n+1})} \tag{1}$$

holds for all $n \geq 1$.

Proof. Plugging $a = (1 + nx_n)^{1/n}$ to the identity $a^n - 1 = (a - 1)\sum_{k=0}^{n-1} a^k$, we get

$$ nx_n = x_{n+1} \sum_{k=0}^{n-1} (1 + nx_n)^{\frac{k}{n}}, \qquad\text{or equivalently,} \qquad \frac{x_n}{x_{n+1}} = \sum_{k=0}^{n-1} (1 + nx_n)^{\frac{k}{n}} \frac{1}{n}. $$

Recognizing the right-hand side as a Riemann sum and noting that $t \mapsto (1 + nx_n)^t$ is increasing, we get

$$ \frac{x_n}{x_{n+1}} \leq \int_{0}^{1} (1 + nx_n)^{t} \, \mathrm{d}t = \psi(nx_n), $$

proving the first half of $\text{(1)}$. Similarly,

\begin{align*} \frac{x_n}{x_{n+1}} = \left( \sum_{k=1}^{n} (1 + nx_n)^{\frac{k}{n}} \frac{1}{n} \right) - x_n \geq \left( \int_{0}^{1} (1 + nx_n)^{t} \, \mathrm{d}t \right) - x_n = \psi(nx) - x_n. \end{align*}

Rearranging this, we obtain the second half of $\text{(1)}$. $\square$

Lemma 3. Let $\psi^{-1} : [1, \infty) \to [0, \infty)$ denote the inverse of $\psi$. Then

$$ \bbox[border:1px green solid;padding:8px;background-color:#e8ffe8;]{2(y-1) \leq \psi^{-1}(y) \leq 2y \log y} \tag{2} $$

for all $y \geq 1$.

Proof. Note that $\psi$ is increasing, and so, it suffices to prove

$$ \psi(2(y-1)) \leq y \leq \psi(2y \log y) $$

for all $ y\geq 1$.

First, by letting $x = 2(y - 1) \geq 0$, the first half of $\text{(2)}$ is equivalent to $\psi(x) \leq 1+x/2$. However, this follows from the Jensen's inequality applied to $ \psi(x) = \bigl( \int_{0}^{1} (1 + xt)^{-1} \, \mathrm{d}t \bigr)^{-1} $.

Next, by letting $z = \log y \geq 0$, the second half of $\text{(2)}$ is equivalent to $e^z \leq \psi(2ze^z)$. It is not hard to check that this is equivalent to $z \leq \sinh z$, which is true. $\square$

Lemma 4. As $n \to \infty$, we have

$$ \bbox[border:1px green solid;padding:8px;background-color:#e8ffe8;]{ x_n \geq \frac{4}{n^2} + \mathcal{O}\left(\frac{1}{n^3}\right).} $$

Proof. From the first half of $\text{(1)}$ and $\text{(2)}$, we get

$$ nx_n \geq \psi^{-1}\left(\frac{x_n}{x_{n+1}}\right) \geq 2\left(\frac{x_n}{x_{n+1}} - 1 \right). ​$$

Dividing both sides by $2x_n$,

$$ \frac{n}{2} \geq \frac{1}{x_{n+1}} - \frac{1}{x_n}. ​$$

Summing both sides for $ 1, 2, \ldots, n-1$, we obtain

$$ \frac{1}{x_n} - \frac{1}{x_1} \leq \sum_{k=1}^{n-1} \frac{k}{2} = \frac{n(n-1)}{4}. $$

From this, it is easy to conclude the desired inequality. $\square$

Lemma 5. As $n\to\infty$, we have

$$ \bbox[border:1px green solid;padding:8px;background-color:#e8ffe8;]{x_n \leq n^{-2+o(1)}.} $$

Proof. From the second half of $\text{(1)}$ and $\text{(2)}$, we get

$$ nx_n \leq \psi^{-1}\left(\frac{x_n}{x_{n+1}}(1 + x_{n+1}) \right) \leq \frac{2x_n}{x_{n+1}}(1 + x_{n+1}) \log \left(\frac{x_n}{x_{n+1}}(1 + x_{n+1}) \right). $$

Rearranging this, we get

$$ n x_{n+1} \left( \frac{1}{2(1 + x_{n+1})} - \frac{1}{n\psi(x_{n+1})} \right) \leq \log x_n - \log x_{n+1}. $$

So by invoking Stolz–Cesàro theorem and applying Lemma 1 and 4 together, we get

\begin{align*} \liminf_{n\to\infty} \frac{-\log x_n}{\log n} &\geq \liminf_{n\to\infty} \frac{\log x_n - \log x_{n+1}}{1/n} \\ &\geq \liminf_{n\to\infty} n^2 x_{n+1} \left( \frac{1}{2(1 + x_{n+1})} - \frac{1}{n\psi(x_{n+1})} \right) \\ &\geq 4 \cdot \frac{1}{2} = 2, \end{align*}

and hence the desired claim follows. $\square$

Now we are ready to answer OP's question.

Corollary 6. We have

$$ \bbox[border:1px green solid;padding:8px;background-color:#e8ffe8;]{\lim_{n\to\infty} n x_n = 0 \qquad\text{and}\qquad \lim_{n\to\infty} \frac{x_{n+1}}{x_n} = 1.} $$

By Lemma 4 and 5 altogether, we have $\frac{4+o(1)}{n^2} \leq x_n \leq \frac{1}{n^{2+o(1)}} $ and hence $n x_n \to 0$. Next, by $\text{(1)}$, we have $1 \leq \frac{x_n}{x_{n+1}} \leq \psi(n x_n)$. So by using the fact that $\psi(x) \to 1$ as $x \to 0$, the second limit follows by the squeezing theorem. $\square$

Finally, we establish:

Theorem 7. We have

$$ \bbox[border:1px green solid;padding:8px;background-color:#e8ffe8;]{\lim_{n\to\infty} n^2 x_n = 4.} $$

Proof. By $\text{(1)}$ and the Taylor expansion $\psi(x) = 1 + \frac{x}{2} + \mathcal{O}(x^2)$ as $x \to 0$, we get

\begin{align*} \frac{x_n}{x_{n+1}} &= \psi(nx_n)(1 + \mathcal{O}(x_{n+1})) \\ &= \left(1 + \frac{nx_n}{2} + \mathcal{O}(n^2 x_n^2) \right)(1 + \mathcal{O}(x_{n+1})). \end{align*}

Dividing both sides by $x_n$ and using $x_{n+1} \leq x_n$,

\begin{align*} \frac{1}{x_{n+1}} &= \left(\frac{1}{x_n} + \frac{n}{2} + \mathcal{O}(n^2 x_n) \right)(1 + \mathcal{O}(x_n)) \\ &= \frac{1}{x_n} + \frac{n}{2} + \mathcal{O}(n^2 x_n) + \mathcal{O}(1) + \mathcal{O}(nx_n) + \mathcal{O}(n^2 x_n^2) \\ &= \frac{1}{x_n} + \frac{n}{2} + n^{o(1)}, \end{align*}

where the last line follows from Lemma 5. So by the Stolz–Cesàro theorem,

$$ \lim_{n\to\infty} \frac{x_n^{-1}}{n^2/4} = \lim_{n\to\infty} \frac{x_{n+1}^{-1} - x_n^{-1}}{n/2} = 1 $$

and the desired claim follows.

Since $\lim_{n\to\infty} x_n=0$, we have that $\lim_{n\to\infty}\frac{x_{n+1}}{x_n}=\lim_{n\to\infty}\frac{-1+\sqrt[n]{1+nx_n}}{x_n}=1$, where we used identity $a-b=\frac{a^n-b^n}{a^{n-1}+\dots+b^{n-1}}$. By using this, it is easy to get by Stolz's theorem that $\lim_{n\to\infty} nx_n=\lim_{n\to\infty}\frac{n+1-n}{\frac{1}{x_{n+1}}-\frac{1}{x_n}}=0$.