Let $(x_n)_{n\geqslant1}$ be a sequence, with $x_1>0$ and $$x_{n+1} = \frac{x_n}{(x_n + 1)^2 + 1}\qquad (1)$$ Study the convergence of the following sequences:$$(x _n)_{n\geqslant1},\ (n^k x _n)_{n\geqslant1},$$ where $k > 0$.
I've thought of doing $\dfrac{x_{n+1}}{x_n}$ so that I obtain that this fraction is smaller than $1$ and therefore the sequence is decreasing as $n$ increases, but I could only find that the sequence is bigger than $0$ by using induction starting from $x_1>1$. Is that enough to pass to the limit and show it is $0$?
For the second task, I think at Stolz-Cesaro's Lemma but I can't figure out how to properly use it as if I transform $(n^kx _n)_{n\geqslant1}$ in $\dfrac{x_n}{1/n^k}$ I obtain something complicated.
Could you show me how to solve it and tell me what to do to complete the first attemp to prove it?
Notice that $x_n>0$ for all $n\in \Bbb N$ since $x_1>0$ therefore:$$(x_n+1)^2+1>2$$ and we have $$|x_{n+1}|={|x_n|\over(x_n+1)^2+1}\le {|x_n|\over 2}\qquad\qquad(*)$$which means that $x_n\to 0$ as $n \to \infty$.
For the second series, there is a very easy proof that $n^kx_n$ always converges to zero. First fix an arbitrary real $a\in (1,2)$ (for example $a={3\over 2}$) and define $y_n=a^n\cdot x_n$ then by multiplying the both sides of $(*)$ in $a^{n+1}$ we obtain$$|a^{n+1}x_{n+1}|\le {|a\cdot a^nx_n|\over2}$$by substituting $y_n$ we have $$|y_{n+1}|\le {a\over 2}|y_n|$$for example with $a=1.5$ we have$$|y_{n+1}|\le 0.75|y_n|$$which means that $y_n=a^n x_n\to 0$. Since $a^n$ for $a>1$ grows super faster than $n^k$ for any real $k$ and $n\in \Bbb N$ we finally conclude that
$n^kx_n$ tends to $0$ for any $k\in \Bbb R$.
Here is an illustration of the convergence for $10\le k\le11$ generated by MATLAB: