Given $n\in \mathbb N$ and: $$ \begin{cases} x_{n+1} = \frac{x_n + 1}{n+1} \\ x_1 = -10 \end{cases} $$ Show that $x_n$ is decreasing starting from some index $n_0$. Find $n_0$.
I've tried to find a closed form of the sequence:
$$ x_{n+1} = \frac{x_n}{n+1} + \frac{1}{n+1}\\ x_n = \frac{x_{n-1}}{n} + \frac{1}{n}\\ x_{n-1} = \frac{x_{n-2}}{n-1} + \frac{1}{n-1}\\ \dots $$
Now consecutively substituting $x_k$ one may obtain: $$ x_{n+1} = x_{n-2}\cdot \frac{1}{(n-1)n(n+1)} + \frac{1}{(n-1)n(n+1)} + \frac{1}{n(n+1)} + \frac{1}{(n+1)} $$
By expanding the pattern we get:
$$ x_{n+1} = \frac{n!}{(n+1)!} + \frac{(n-1)!}{(n+1)!} +\dots \frac{1}{(n+1)!} + x_1\cdot\frac{1}{(n+1)!} = \left(\sum_{k=1}^n\frac{(n+1-k)!}{(n+1)!}\right) + \frac{x_1}{(n+1)!} $$
I couldn't find a closed form for the sum.
I've been also thinking of another way to show that. If we assume that $x_n$ is indeed decreasing starting from some $x_k$ then there should exist a supremum of the sequence and the following should hold:
$$ \begin{cases} x_k > x_{k-1} \\ x_k > x_{k+1} \end{cases} $$
But i'm not sure how to isolate $k$ from that system. I've also tried to play around with the consecutive terms in order to arrive at some linear recurrence and try to find its closed form by utilizing its characteristic polynomial but failed.
Could you please show me how to prove the problem statement and find the required $n_0$
Please note the precalculus tag. Thank you!
Solve
$$\frac{x+1}{n+1}<x,$$ which gives
$$nx>1.$$
Then the first few terms are
$$-10,-\frac92,-\frac76,-\frac1{24},\frac{23}{120},\frac{143}{720}$$ and the last one exceeds $\dfrac1n$.