How to find the limit of $x_n$ if $x_1=1; x_{n+1}=2+\frac {3}{x_n}+\frac{1}{n}$. How to find the limit of $x_n$ if $x_1=1; x_{n+1}=2+\frac {3}{x_n}+\frac{1}{n}$.
My idea is to limit $x_n$ from both sides with something, and prove that these sequences' limits are the same.
So, I can say that:
$y_{n+1} = 2 + \frac {3}{x_n} \leq 2+\frac {3}{x_n}+\frac{1}{n} = x_{n+1}$. Аnd its limit is 3.
But I cannot find any sequence $z_n$ whose values ​​satisfy the inequality $x_n \leq z_n$ and $\lim { z_n } = 3$.
Can someone help me with this task, please. Maybe exist any other solution of this task. I have no more ideas so that's why I'm here.

If you can show the sequence is bounded (below, as Calvin's hint suggests) and eventually monotonic (decreasing), then let $L$ be the limit of the sequence, so that
$$\lim_{n\to\infty}x_{n+1}=\lim_{n\to\infty}\left(2+\frac3{x_n}+\frac1n\right)\implies L=2+\frac3L$$
Then solve for $L$.
