<s>[INST] Prove the positivity of a sequence Let $a>0$ a real number and $(u_n)$ the sequence defined by 
$$
u_{n+1} = a - \frac{1}{u_n}\text{ and } u_0 = a.
$$
Question: Determine condition on the value of $a>0$ such that the sequence $(u_n)$ is always positive. 
Attempt: I tried to establish a general formula of $u_n$ in order to set up a condition on $a$, by calculating $u_n$ with some $n$:
$$
u_1 = a - \frac{1}{u_0} = a - \frac 1a = \frac{a^2-1}{a},\\
u_2 = a - \frac{1}{u_1}= a - \frac{a}{a^2-1}=\frac{a^3-2a}{a^2-1},\\
u_3 = a - \frac{1}{u_2} = a - \frac{a^2-1}{a^3-2a}=\frac{a^4-3a^2+1}{a^3-2a},\\
u_4 = a - \frac{1}{u_3} = a - \frac{a^3-2a}{a^4-3a^2+1}=\frac{a^5-4a^3+2a-1}{a^4-3a^2+1}.
$$
But I wasn't successful, it seems that there is no general formula of $u_n$. So I would be appreciate for any suggestion of solution. Thank you!
 [/INST] Notice that if the sequence $u_n$ is positive, then it must be decreasing. This can be proven by induction.
In this case, the sequence $u_n$ is monotonic and bounded, so it has a limit $L$ that has to be positive.
From the formula,
$$L=a-\frac{1}{L}$$
Thus
$$a=L+\frac{1}{L} \ge 2$$
So a necessary condition is $a \ge 2$.
To prove that the condition is suffecient, Take $L>0$ such that $a=L+\frac{1}{L}$ and prove by induction that $u_n$ is decreasing, positive and $u_n > L$.
</s>