Find the limit of $x_n$ if $x_1=1; x_{n+1}=2+\frac {3}{x_n}+\frac{1}{n}$.

Question

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.

Answer 1

Hint: Set up a double sided inequality to bound $x_{n+1}$ for $n$ large enough. This can be proved via induction.

$ 3 < x_n < 3 + \frac{1}{n-1}$.

Hence conclude that $\lim x_n = 3 $.