Finding a limit of a recurrently given sequence

Question

I am struggling with one example of limit required for recurrently given sequence. There are 3 examples in total, I will also post the ones I did.

$ x_1>0$ and recursively given sequence: $$x_{n+1}=\frac{x_n}{1+x_n+x_n^2}$$

Find the following limits: $\displaystyle \lim_{n \to \infty }x_n;$ $ \displaystyle \lim_{n \to \infty }nx_n;$ and $\displaystyle \lim_{n \to \infty }\frac{n(1-nx_n)}{\log_{e}n}$

First thing I do is determining whether or not the sequence is monotonically increasing or decreasing. (I already know the limit of this sequence will be $\geq 0$ because of $x_1>0$)

$\frac{x_{n+1}}{x_n}$ gives me $\frac{1}{1+x_n+x_n^2}$ which is $<1$ so the sequence monotonically decreases

I take some $L= \displaystyle \lim_{n \to \infty }x_n = \displaystyle \lim_{n \to \infty }x_{n+1}$. Swapping this into the starting sequence I get $L=\frac{L}{1+L+L^2}$ and from here I get $L=0 \vee L=-1$

Because of $x_1>0$, I already know $\displaystyle \lim_{n \to \infty }x_n \neq -1$, so $\displaystyle \lim_{n \to \infty }x_n = 0$ is the only solution here.

The next example is also pretty straight forward, I take $x_n=\frac{y_n}{n}$ and $x_{n+1}=\frac{y_{n+1}}{n+1}$. Here, the result is $\displaystyle \lim_{n \to \infty }nx_n=1$.

The third example, $\displaystyle \lim_{n \to \infty }\frac{n(1-nx_n)}{\log_{e}n}$, looks the trickiest and I couldn't solve it using the method I used in the 2 examples above. If anybody has any idea about solving it please share it. By the way, I am not sure if my method of solving is actually good, so if it's not please recommend/teach me a more versatile method.

Answer 1

Let $y_n=\frac1{x_n}$, then the condition $$x_{n+1}=\frac{x_n}{1+x_n+x_n^2}$$ transforms to $$y_{n+1}=y_n+1+\frac1{y_n}.$$ (1) $y_{n+1}-y_n>1$ implies $\{y_n\}$ is increasing and $y_n=\sum_{k=1}^{n-1}(y_k-y_{k-1})>n-1$ is unboubded above. So $$\lim_{n\to\infty}y_n=\infty\implies\lim_{n\to\infty}x_n=0.$$ (2) By Stolz's theorem, $$\lim_{n\to\infty}\frac{n}{y_n} =\lim_{n\to\infty}\frac{(n+1)-n}{y_{n+1}-y_n} =\lim_{n\to\infty}\frac{1}{1+\frac1{y_n}} =1\implies\lim_{n\to\infty}nx_n=1$$

(3) By Stolz''s theorem again (for detailed computation, see the comment below), $$\lim_{n\to\infty}\frac{y_n-n}{\log n} =\lim_{n\to\infty}\frac{\frac1{y_n}}{\log(1+1/n)} =\lim_{n\to\infty}\frac{\frac{n}{y_n}}{n\log(1+1/n)} =1.$$ Then, for the thrid limit, $$\lim_{n\to\infty}\frac{n(1-nx_n)}{\log n} =\lim_{n\to\infty}\frac{n\left(1-\frac{n}{y_n}\right)}{\log n} =\lim_{n\to\infty}\frac{n(y_n-n)}{y_n\log n} =\lim_{n\to\infty}\left[\frac{n}{y_n}\cdot\frac{y_n-n}{\log n}\right] =1.$$